Download Problem 5 - grandpasfsc105

KINEMATICS
1.
A box sits on a horizontal wooden board. The coefficient of static friction
between the box and the board is 0.5. You grab one end of the board and lift it up,
keeping the other end of the board on the ground. What is the angle between the
board and the horizontal direction when the box begins to slide down the board?
Solution:
The critical angle is determined by the condition:
From this equation we can find an angle 26 0
2.
An archer shoots an arrow with a velocity of 30 m/s at an angle of 20 degrees
with respect to the horizontal. An assistant standing on the level ground 30 m
downrange from the launch point throws an apple straight up with the minimum
initial speed necessary to meet the path of the arrow. What is the initial speed of
the apple and at what time after the arrow is launched should the apple be thrown
so that the arrow hits the apple?
Solution:
The motion of the arrow is a projectile motion. Then the motion of the arrow
along horizontal direction is a motion with constant velocity:
Where the initial position is 0 and
. Then
Motion along vertical axis is a motion with constant acceleration, then
Where
, initial position is 0, and
. Then
Then we need to find the time when the arrow will be exactly above the assistant.
At this moment of time x(t) = 30. Then we can find the time:
Then we can find the height of the arrow at this moment of time:
The assistant should throw the apple with minimal velocity so it will reach point
5.4 m and at this height the velocity should be 0. From this condition we can find
the minimal velocity:
The time of the motion of the apple to this point is
Then the apple should be thrown after
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3.
An 8 kg block is at rest on a horizontal floor. If you push horizontally on the 8 kg
block with a force of 20 N, it just starts to move.
(a) What is the coefficient of static friction?
(b) A 10.0 kg block is stacked on top of the 8 kg block. What is the magnitude F
of the force, acting horizontally on the 8 kg block as before, that is required to
make the two blocks start to move?
Solution:
The magnitude of horizontal force should be equal to the magnitude of the
maximal static friction force, which is equal to the product of the coefficient of
static friction and the normal force (gravitation force in the present problem).
(a) The gravitation force is mg=8*9.8 = 78.4 N. Then the coefficient of static
friction is
(b) Now we know the coefficient of static friction and we know the normal force:
18*9.8 = 176.4 N. Then we can find the magnitude of force F:
4.
You are driving along the street at the speed limit (35mph) and 50 meters before
reaching a traffic light you notice it becoming yellow. You accelerate to make the
traffic light within the 3 seconds it takes for it to turn red. What is your speed as
you cross the intersection? Assume that the acceleration is constant and that there
is no air resistance.
Solution:
This is the motion with constant acceleration. If the acceleration of the car is a
then we can write the expression for traveled distance:
We know the initial velocity
.
We also know that after 3 seconds the car travels distance 50 meters. Then we can
find acceleration
Now we know acceleration, then we can find the final velocity:
2
5.
The car drives straight off the edge of a cliff that is 57 m high. The investigator at
the scene of the accident notes that the point of impact is 130 m from the base of
the cliff. How fast was the car traveling when it went over the cliff?
Solution:
Motion along axis x (horizontal axis) is the motion with constant velocity. So we
can write down the dependence of x-coordinate as a function of time:
Where v is the initial velocity (the initial velocity has only x-component, its
direction is along axis x).
We know the final x-coordinate of the car – it is 130 m. But we do not know the
traveled time and the initial velocity.
We can find the traveled time from the motion along axis y. This is the motion
with constant acceleration (free fall acceleration). We know the height of the cliff
– this is the traveled distance in y-direction. We know that initial velocity (in ydirection) is 0. Then we can write the following equation:
From this equation we can find the traveled time:
Then from the motion along axis x we have:
From this equation we can find initial velocity:
6.
A tortoise and a hare are in a road race to defend the honor of their breed. The
tortoise crawls the entire 1000 meters at a speed of 0.2 m/s. The rabbit runs the
first 200 meters at 2 m/s, stops to take a nap for 1.3 hours, and awakens to finish
the last 800 meters with an average speed of 3 m/s. Who wins the race and by
how much time?
Solution:
At first let us calculate the travel time of tortoise. We know the speed and the
distance, so we can easily find the time:
Now let us calculate the traveled time of rabbit. The traveled time consists of
three parts:
1. He runs the first 200 m at 2 m/s. The time of this motion is
3
2. Then he takes a nap for 1.3 hours:
3. Then he runs the last 800 m with speed 3 m/s. The time of this motion is
Then the total travel time is
Since t p  t r then tortoise wins the race by 47 s.
7.
A "moving sidewalk" in a busy airport terminal moves 1 m/s and is 200 m long. A
passenger steps onto one end and walks, in the same direction as the sidewalk is
moving, at a rate of 2.0 m/s relative to the moving sidewalk. How much time does
it take the passenger to reach the opposite end of the walkway?
Solution:
The speed of the passenger relative to the ground is 2m/s+1m/s =3 m/s (since the
he is walking in the same direction as the direction of the motion of the sidewalk.
Then the passenger reaches the end of the sidewalk after
8.
(a) If a particle's position is given by x  6  12t  4t 2 (where t is in seconds, and x
is in meters), what is its velocity at t = 1s?
(b) What is its speed at t=1s?
(c) Is there ever an instant when the velocity is 0? If so, give the time.
Solution:
The velocity is the derivative of x(t) with respect to time. Then
(a) at t = 1 s we get:
What we calculate here is the x-component of velocity. The negative sign means
that the direction of velocity is opposite to the direction of axis x.
(b) the speed is the magnitude of velocity. Then at t = 1 s the speed is 4 m/s
(c) to find time at which the velocity is 0 we just need to solve an equation:
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From this equation we find time:
At this moment of time the velocity is 0.
8.
The captain of a plane wishes to proceed due west. The cruising speed of the
plane is 245 m/s relative to the air. A weather report indicates that a 38-m/s wind
is blowing from the south to the north. In what direction, measured due west,
should the pilot head the plane relative to the air?
Solution:
This is problem is on relative motion. The velocity of the plane is the vector sum
of the velocity of the wind (air) and the relative velocity of the plane (relative to
the air) as shown in the figure.
We know that the direction of the final velocity of the plane is from east to west
(as shown in the figure).
Then from the triangle base shown in the figure we can find angle  . This angle
characterizes the direction of the relative velocity of the plane.
Then
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9.
The highest barrier that a projectile can clear is 14 m when the projectile is
launched at an angle of 30.0 degrees above the horizontal. What is the projectile's
launch speed?
Solution:
The maximum height of the projectile is given by the equation:
where   30 0 is the launch angle. Then we can find the initial velocity:
10.
Snow is falling vertically at a constant speed of 3 m/s. At what angle from the
vertical do the snowflakes appear to be falling as viewed by the driver of a car
traveling on a straight, level road with a speed of 60 km/h?
Solution:
At first we need to have the same units, so we need to convert the speed of the car
in m/s:
Then we need to find the direction of the velocity of the snowflakes relative to the
car:
Then from the triangle shown in the picture we can find angle :
Then
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11.
A missile is launched directly upward into the air at an initial velocity of 80 m/s.
It is moving with constant velocity until it reaches 1000m, when the engine fails.
(a) How long does it take it to reach 1000m?
(b) How high does the missile go?
(c) How long does it take for it to fall back to the earth?
(d) How long does it stay in the air?
(e) How fast is it going when it hits the ground?
Solution:
(a) Since initially the motion is with constant velocity, we can easily find the time
of the motion of the missile till it reaches the height 1000 m. The time is given by
the expression:
After this point there is free fall motion – there is only one force acting on the
object (gravitational force) – this force provides free fall acceleration 9.8 m / s 2 .
The initial velocity is 80 m/s pointing upward. The acceleration is pointing
downward. The initial height of the missile is 1000 m. Then the equations which
describe this motion are the following:
(b) To find the maximum height of the missile we can use the last equation. The
velocity at the maximum height is 0. Then
(c) To find the time when the missile hits the ground we need to use the first
equation:
When the missile hits the ground h=0. Then
From this equation we can find time: 24.6 s.
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(d) Then we can find the time when the missile is in the air: it is the sum of the
time when it reaches 1000 m and the time when it hits the ground:
(e) To find the speed of the missile when it hits the ground we need to use the last
equation:
When the missile hits the ground h = 0. Then
v  80 2  2 * 9.8 *1000 = 161 m/ s
Here we need to add the minus sign, which illustrate the fact that the direction of
velocity is downward.
12.
A skier is accelerating down a 30 degree hill at 3 m / s 2 .
(a) What is the vertical component of her acceleration?
(b) How long will it take her to reach the bottom of the hill, assuming she starts
from rest and accelerates uniformly, if the elevation is 300 m?
Solution:
(a) If we assume that the direction of the vertical axis is downward then the
vertical component is positive and is equal to
(b) This is the motion with constant acceleration then the equation which describe
this motion is the following:
The traveled distance (L) is related to the elevation (h) by the equation:
Now we know acceleration and the traveled distance then we can find the traveled
time:
13.
A rocket is fired vertically upwards with initial velocity 80 m/s at the ground
level. Its engines then fire and it is accelerated at 4 m / s 2 until it reaches an
altitude of 1000 m. At that point the engines fail and the rocket goes into free-fall.
Disregard air resistance.
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(a) How long was the rocket above the ground?
(b) What is the maximum altitude?
(c) What is the velocity just before it collides with the ground?
Solution:
The first part of the motion is the motion with constant acceleration at
.
The initial velocity for this motion is 80 m/s. Then we can write the equation,
which describe the dependence of height of the rocket on time:
From this equation we can find the time when the rocket reach the height 1000 m
= h:
The solution of this equation is 10 s. So after 10 seconds the engine fails. The
velocity at this moment of time is
After this moment of time we have free fall motion – there is only one force
acting on the object (it is gravitational force) – this force provide free fall
acceleration.
The initial velocity is 120 m/s pointing upward. The acceleration is
pointing downward. The initial height of the rocket is 1000 m. Then the
equations which describe this motion are the following:
To find the maximum height of the rocket we can use the last equation. The
velocity at the maximum height is 0. Then
This is the answer to part (b).
To find the time when the rocket hits the ground we need to use the first equation:
When the rocket hits the ground h=0. Then
From this equation we can find time: 31 s.
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Then we can find the time when the rocket is in the air: it is the sum of the time
when it reaches 1000 m and the time when it hits the ground:
This is the answer to part (a).
To find the speed of the rocket when it hits the ground we need to use the last
equation:
When the missile hits the ground h=0. Then
This is the answer to part (c).
14.
A rock is dropped from rest into a well. The sound of the splash is heard 4 s after
the rock is released from rest.
(a) How far below to top of the well is the surface of the water? (the speed of
sound in air at ambient temperature is 336m/s).
(b) If the travel time for the sound is neglected, what % error is introduced when
the depth of the well is calculated?
Solution:
(a) If h is the depth of the well then to find the time of the rock's fall we need to
use the following equation:
From this equation we can find time:
After the rock hits the water the sound of splash will propagate the distance h with
h
speed 336 m/s. After time t s 
, the sound reaches the ground. Then the total
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time taken is
This time is equal to 4 s. Then
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From this equation we can find h:
(b) If we neglect the sound velocity then t = 4 s and we can find the height from
the equation:
If we compare this result with the result from part (a) then we can find an error:
15.
A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the
bull's-eye on a dartboard. It hits at point Q on the rim, vertically below P, 0.2 s
later. What is the distance PQ? How far away from the dartboard is the dart
released?
Solution:
The initial velocity is horizontal; the magnitude is 10 m/s. Then the equation of
motion along horizontal axis has the following form:
After 0.2 seconds the dart hits the target. From this condition and from the above
equation we can find the distance between the point where the dart is released and
the board:
The equations which describe the motion along the vertical axis are the following:
If we substitute into the first equation the traveled time then we can find the
distance between points P and Q:
The minus sign means that point Q is below point P.
16.
A swimmer dived off a cliff with a running horizontal leap. What must his
minimum speed be just as he leaves the top of the cliff so that he will miss the
ledge at the bottom which is 2 m wide and 9 m below the top of the cliff?
Solution:
This is the projectile motion with horizontal initial velocity. In this case the
motion in horizontal direction is described by the following equation:
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where v is the initial velocity.
The motion along vertical axis y is described by the following equations (ycomponent of initial velocity is 0):
We know that the final point of the motion is the end of the ledge. The
coordinates of the final point are: x=2 m and y = -9m.
We can substitute these numbers in the above equations and obtain:
From the seconds equation we can find time:
Then we substitute this time in the first equation and obtain the initial velocity:
17.
A ferris wheel with radius 20 m which rotates counterclockwise, is just starting
up. At a given moment, a passenger on the rim of the wheel and passing through
the lowest point of his circular motion is moving at 3.00m/s and in gaining speed
at a rate of 0.5 m / s 2 . Find the magnitude of the passenger's acceleration at the
instant.
Solution:
The acceleration at this point has two components: the tangential acceleration,
which is equal to at = 0.5 m / s 2 , and centripetal acceleration, which is given by
the expression:
These components are orthogonal. Then the net acceleration is
18.
In an action film, the hero is supposed to throw a grenade from his car, which is
going at 90 km/h, at his enemy's car, which is going 110 km/h. The enemy's car is
15.8 m in front of the hero's when he lets go of the grenade. If the hero throws the
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grenade so its initial velocity relative to him is at an angle of 45(degree) above the
horizontal, what should the magnitude of the initial velocity be? The cars are both
traveling in the same direction on a level road. Ignore air resistance.
Find the magnitude of the velocity both relative to the hero and relative to the
earth.
Solution:
At first we need to convert all the velocities into m/s:
We will describe the motion in the hero's reference frame. In this reference frame
the velocity of the enemy's car is
And the grenade is released with the launch angle 450. We introduce the
magnitude of initial velocity of grenade (relative to the hero's car) as v.
After the grenade is released it is moving according to the equations of projectile
motion. Along axis x there is motion with constant velocity:
Along axis y there is motion with constant acceleration:
In these equations at the initial moment of time the grenade has zero coordinates.
We can also write down the equation of enemy's car motion:
Now we need to write down the condition that the grenade hits the enemy's car:
At some moment of time t 0 the x coordinate of grenade should be equation to the
x-coordinate of enemy's car and at the same moment of time the y-coordinate of
grenade should be equal 0.
Then we have:
Then we just need to solve the system of two equations with two unknown
variables:
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Then
And
Then
This is the velocity of the grenade relative to the hero.
We can also find the magnitude of velocity relative to the earth. The x-component
of this velocity is
The y-component of the velocity is
Then the magnitude is
This is the magnitude of the velocity of grenade relative to the earth.
19.
A flowerpot falls from a windowsill 25.0 m above the sidewalk.
(a) how fast is the flowerpot moving when it strikes the ground?
(b) how much time does a passerby on the sidewalk below have to move out of
the way before the flowerpot hits the ground?
Solution:
This is the free fall motion with constant acceleration. The equations which
describe this motion are the following:
where we introduced an axis y with upward direction. The origin of axis y is the
initial position of the flowerpot. The initial velocity of flowerpot is 0.
We know the final position of the flowerpot: the coordinate of the final point is 25 m.
(a) We substitute this coordinate in the third equation an obtain the velocity:
The negative sign means that the direction of velocity is downward. The
magnitude of velocity is 22 m/s.
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(b) The traveled time can be found from the first equation in the above system.
We substitute y(t) = -25 and obtain time:
20.
A firefighter, a distance d from a burning building, directs a stream of water from
a fire hose at angle  above the horizontal. If the initial speed of the stream is v ,
at what height h does the water strike the building?
Solution:
This is projectile motion. The equation which describe the motion along
horizontal axis x is the following
Along axis y we have motion with constant acceleration:
In these equations at the initial moment of time the water has zero coordinate.
What do we know about the final point: we know only the x-coordinate of the
final point – it is x = d. We can substitute this coordinate in the first equation
(which describe the motion along axis x):
From this equation we can find the traveled time:
Then we substitute this time into the equation which describe the y-coordinate of
the water:
This expression gives as the height where the water strikes the building.
21.
You are driving towards a traffic signal when it turns yellow. Your speed is 55
km/h, and your best deceleration has the magnitude of a  5.15 m / s 2 . Your best
reaction time before braking is t 0  0.75 s . To avoid having the front of your car
enter the intersection after the light turns red, should you break to a stop or
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continue to move at 55 km/h if the distance to the intersection and the duration of
the yellow light are
(a) 40m and 2.8s, and
(b) 32m and 1.8s?
Solution:
The first step – we need to convert the speed to m/s:
We have two possible types of motion:
1. motion with constant speed till the driver reaches the intersection. Time of this
motion is
2. motion with deceleration. The driver starts deceleration after t 0  0.75 s .
During this time the driver travels the distance:
The equations which describe these motion are the following:
where the initial position of the car has zero coordinate.
At the final point the velocity is 0. Then
During this time the position of the car is
Now we can analyze the problem:
(a) 40m and 2.8s. Then for motion 1 we have:
Since 2.6<2.8 then the driver can cross the intersection if he will move with
constant speed.
(b) 32m and 1.8s. In this case:
Since 2.1>1.8 the driver cannot cross the intersection.
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Therefore in this part we need to consider case 2. But even in this case the
distance he will travel before he stops is 34.2 m, which is greater than 32 m.
Therefore he cannot stop before the intersection.
22.
The tips of the blades in a food blender are moving with a speed of 20 m/s in a
circle that has a radius of 0.06 m. How much time does it take for the blades to
make one revolution?
Solution:
The traveled distance of a tip is
where R = 0.06 m.
Then the time is equal to the ratio of the traveled time and the average velocity:
23.
While you are traveling in a car on a straight road at 90 km/hr, another car passes
you in the same direction; its speedometer reads 120km/hr. What is your velocity
relative to the other driver? What is the other car's velocity relative to you?
Solution:
The relative velocity is equal to the difference of velocities.
So if you need to find your relative velocity then you need to subtract from your
velocity the velocity of another car:
The minus sign here gives us the direction of relative velocity: the direction is
opposite to the directions of motions of the cars.
Similarly we can find:
The plus sign here gives us the direction of relative velocity: the direction is the
same as the directions of motions of the cars.
24.
A tortoise can run with a speed of 10.0 cm/s, and a hare can run exactly 10 times
as fast. In a race, they both start at the same time, but the hare stops to rest for
3.00 min. The tortoise wins by 10 cm.
(a) How long does the race take?
(b) What is the length of the race?
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Solution:
(a) If L is the length of the race then the tortoise run this race the time:
When the tortoise crosses the finish line the hare is 20 cm = 0.2 m behind tortoise.
It means that during time t tortoise it traveled the distance L - 0.2. Since he stopped
for rest for 3 minutes =3*60 s=180 s and his speed is 10vtortoise  1m / s , then we
can write the equation:
Taking into account the first equation we have:
Then
(b) Then we can find the length of the race:
25.
Emily takes a trip, driving with a constant velocity of 90 km/h to the north except
for a 30 min rest stop. If Emily's average velocity is 75 km/h to the north, how
long does the trip take?
Solution:
The average velocity is the ratio of traveled distance and the traveled time:
If the traveled time is t, then Emily traveled time (t-0.5)h with the speed 90 km/h.
Then the traveled distance is
We substitute this relation in the first equation and obtain:
Then
Then we can find L (length of the trip):
26.
To qualify for the finals in a racing event, a racecar must achieve an average
speed of 250 km/h on a track with a total length of 2000 m. If a particular car
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covers the first half of the track at an average speed of 230 km/h, what minimum
average speed must it have in the second half of the event to qualify?
Solution:
The average speed is the ratio of traveled distance and traveled time:
For the whole track we have: vav  250 km / h and L  2000 m  2 km . Then the
traveled time is
The car travels the first half of the track (1 km) with speed 230 km/h. The time of
this motion is
Since the total traveled time should be 0.008 h, then the second part of the track
the car should travel 0.008-0.0043 = 0.0037 h. Since the distance of this motion is
1 km, then the average speed is
27.
You are trying to cross a river that flows due south with a strong current. You
start out in your motorboat on the west bank, desiring to reach the east bank
directly across from your starting point. Which direction should head your
motorboat? Draw a picture of the river, the banks, and your motorboat, and
include the relevant velocity vectors. What information would you need in order
to determine the actual direction you need to head?
Solution:
To find the actual direction of the boat velocity we need to know the velocity of
the river and the speed of the boat relative to the river (relative velocity). Then the
actual velocity (this is the vector) is
The diagram is shown below.
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28.
The acceleration of an object as a function of time is a(t )  8t (m / s 2 ) .
Determine the
(a) velocity and
(b) the position of the object as a function of time if it is located at x = 2 m and
has a velocity of 3 m/s at time t = 0 s.
Solution:
(a) By definition, the velocity of the object is the integral of acceleration with
respect to time:
where v0  3 m / s is the initial velocity (at t = 0).
(b) The position can found as an integral of velocity with respect to time:
where x0  2 m is the initial position (at t = 0) of the object.
29.
An automobile traveling 90 km/h overtakes a 1.5-km-long train traveling in the
same direction on a track parallel to the road. If the train's speed is 70 km/h,
(a) how long does it take the car to pass it, and
(b) how far will the car have traveled in the time?
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Solution:
It is easier to solve this problem if we introduce the relative velocity of the car
(relative to the train). The velocity of the car relative to the train is (90-70) km/h =
20 km/h.
In the relative description the train is not moving and the car is moving with
constant speed 20 km/h.
(a) Then we can easily find the time the car needs to pass the train. It is
(b) To find the actual traveled distance of the car we just need to multiply the
traveled time (0.075 h) by the actual speed of the car:
30.
Two cannonballs, A and B, are fired from the ground with identical initial speeds,
but with launch angle of cannonball A larger than the launch angle of cannonball
B.
(a) Which cannonball reaches a higher elevation?
(b) Which cannonball stays longer in the air?
(c) Which cannonball travels farther?
Solution:
(a) The vertical component of the initial velocity of the projectile is proportional
to sine of the launch angle. It means that the vertical component of initial velocity
of cannonball A is larger than the vertical component of cannonball B. Then
cannonball A reaches a higher elevation.
(b) The time in the air depends only on the vertical component of initial velocity
of projectile. The larger the vertical component of the velocity the larger the time
in the air. Then the cannonball A will stay longer in the air.
(c) From the data of the problem we cannot tell which one will travel farther. The
horizontal length is proportional to sin( 2 * launch angle)) . It has maximum at
angle 450. Therefore we can have both possibilities:
cannonball A travels further than cannonball B for example, if the launch angle of
cannonball A is 450 and the launch angle of cannonball B is 300.
Cannonball B travels further than cannonball A for example, if the launch angle
of cannonball B is 450 and launch angler of cannonball A is 500.
31.
A man is driving at the speed 40 mph when he sees an obstacle at distance 300 ft
ahead of his position. The driver applies the brakes and decelerates at 10 ft / s 2 .
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How long does it take him to stop the vehicle? How far from the obstacle will the
driver be when he finally stops?
Solution:
At first we need to convert all the variables into correct units (SI units):
Then we have motion with constant deceleration. Then
The car stops when v(t )  0 , then
And
The traveled distance is
Then the distance between the car and the obstacle is
32.
A body moves 4 km towards East from a fixed point A and reaches point B. Then
it covers 5 km towards North and arrives at point C. Find the distance and
directions of the net displacement.
Solution:
We show two displacements (travelled paths) in the figure.
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The net (resultant) displacement is the vector sum of two displacements: the first


displacement ( s1 ) and the second displacement ( s2 ):
The easiest way to find the net displacement is to introduce coordinate system
(axes x and y) and then find the x and y components of the net vector
displacement. The x and y components of the displacement ( s1 ) and displacement

( s2 ) are the following:
Then the x and y components of the net displacement is
Then the magnitude of the net displacement is
The direction of the net displacement is characterized by angle  (shown in the
figure), which can be found from the known x and y components of the net
displacement:
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33.
A baseball player hits a homerun, and the ball lands in the left field seats, which is
120 m away from the point at which the ball was hit. The ball lands with a
velocity of 20 m/s at an angle of 30 degrees below horizontal. Ignoring air
resistance
(A) find the initial velocity and the angle above horizontal with which the ball
leaves the bat;
(B) find the height of the ball relatively to the ground.
Solution:
(A) Initial velocity
Without air resistance this is simple projectile motion. In the present problem we
do not know initial velocity: we do not know the magnitude of the velocity
(speed) and we do not know its direction.
There are two sets of equations, which describe the motion of the projectile (ball).
Set 1: motion along horizontal axis (axis x – see figure). This is the motion with
constant velocity. There is only one equation, which describe this motion:
................................................(1)
Here x0  0 .
Since the motion along the axis x is the motion with constant velocity then the xcomponent of the velocity is constant. We know the velocity at the final point.
Then we can find the x-component of the velocity at the final point:
This x-component of the velocity is equal to the x-component of the initial
velocity:
...........(2)
We also know the x-coordinate of the final point (point B): it is 120 m. We
substitute this value in equation (1) and obtain
From this equation we can find the time of travel from point A to point B:
Now we need to analyze the second set of equations.
Set 2: motion along vertical axis (axis y – see figure). This is the motion with
constant acceleration – free fall motion. There are three equations, which describe
this motion. Only two equations are independent, but it is convenient to write all
three equations:
24
.............(3)
................................................(4)
Since the initial y-coordinate is zero, then
.............................................(5)
We know the y component of the final velocity
This is the y-component of the velocity at the moment of time t AB  6.9 s . We
substitute these values in equation (4) and obtain
From this equation we can find the y-component of the initial velocity:
Finally we know the x- and y-components of the initial velocity:
25
From these expressions we can find the magnitude of the initial velocity and the
direction (angle) of the initial velocity:
Now we know the initial velocity.
(B) Final height
We need to find the final height of the ball (the final y-coordinate). To find the
final height we can use equation (3). We just need to substitute the y-component
of the initial velocity and the traveled time in this equation:
26
DYNAMICS
34.
You are driving along an empty straight road at a constant speed u. At some point
you notice a tall wall at a distance D in front of you. Would it require a larger
force to (a) continue moving straight and decelerate to a full stop before the wall,
or (b) turn left or right to avoid the wall? (to make the calculation easier assume
that the turn is done at a constant speed along a circular path).
Solution:
(a) If the car continue moving straight then the acceleration should satisfy the
following equation:
where v f , the final velocity, is 0, since the car should stop just before the wall.
vi , the initial velocity, is equal to u. Then
If the mass of the car is m, then the force required to stop the car is
(b) Now the car is turning left or right to avoid the wall. This is the uniform
circular motion with a radius D. The acceleration of this motion is
Then the force required to turn the car is
We can see that in the case (a) the force is 2 times smaller.
35.
In the film 2001: A Space Odyssey, a wheel-like space station achieves artificial
gravity by spinning around its axis. If the station had a size of 2 km, how fast
should it be spinning for the people inside to feel the same gravitational
acceleration as on earth?
Solution:
If the space station is rotating with angular velocity then the acceleration
(centripetal acceleration) at distance R=2 km =2000 m is given by the equation
This acceleration should be equal to the free fall acceleration. Then
27
and
36.
As part a of the drawing shows, two blocks are connected by a rope that passes
over a set of pulleys. The block 1 has a weight of 400 N, and the block 2 has a
weight of 600 N. The rope and the pulleys are massless and there is no friction.
(a) What is the acceleration of the lighter block?
(b) Suppose that the heavier block is removed, and a downward force of 600 N is
provided by someone pulling on the rope, as part b of the drawing shows. Find the
acceleration of the remaining block.
Solution:
(a) The acceleration of the lighter block is equal by magnitude to the acceleration
of the other block. If the tension of the rope is T, then the equations of motion for
block 1 and block 2 are the following:
Then
and
28
(b) In this case the tension of the rope is given: T = 800N. Then the second
Newton's law for block 1 becomes:
37.
A boy of mass 40 kg wishes to play on pivoted seesaw with his dog of mass 15
kg. When the dog sits at 3 m from the pivot, where must the boy sit if the 6.5 m
long board is to be balanced horizontally?
Solution:
We have equilibrium. It means that the net torque should be equal to 0. If x boy is
the distance from the boy to the pivot, then the equilibrium condition becomes:
38.
How fast should the earth spin in order for a 150 lb human not to be able to walk
on the ground?
Solution:
The condition that the human cannot walk on the ground is that there is no contact
between human and the ground. It means that the normal force is 0.
Let us assume that the earth is rotating with angular velocity  . Then the human
on the ground will have centripetal acceleration with the magnitude
Where R  6.4 *10 6 m is the radius of the earth.
This acceleration is provided by the gravitation force and the normal force, so that
Then
Then condition that the normal force is 0 (or negative) is
Then
and
29
39.
Part a of the drawing shows a block suspended from the pulley; the tension in the
rope is 80 N. Part b shows the same block being pulled up at a constant velocity.
What is the tension in the rope in part b?
Solution:
In part (a) the equilibrium condition can be written as:
In part (b) we have the motion with constant velocity. It means that the
acceleration is 0 and the net force is 0, which can be written as
From these two equations we can find
40.
A box is sliding up an incline that makes an angle of 20 degrees with respect to
the horizontal. The coefficient of kinetic friction between the box and the surface
of the incline is 0.2. The initial speed of the box at the bottom of the incline is 2
m/s. How far does the box travel along the incline before coming to rest?
Solution:
The first part in the problem is to find an acceleration of the motion. The
acceleration is due to gravitation force and the friction force and has the following
form:
30
The second part is to write down the kinematic equations of motion. In this
problem we need to use the relation between the traveled distance and initial and
final (the final velocity is 0) velocities:
where s is the traveled distance. Then
41.
A block weighing 80 N rests on a plane inclined at 30 degrees to the horizontal.
The coefficients of static and kinetic friction are 0.2 and 0.1 respectively. What is
the minimum magnitude of the force F, parallel to the plane that will prevent the
block from slipping?
Solution:
The minimum force corresponds to the condition that the static friction force has
the maximum value, which is 0.2*normal force. To find the normal force and the
external force we need to write down the condition of equilibrium: the net force is
0. Then we rewrite this equation in terms of x and y-components (x axis is parallel
to the plane).
The x-component of the second Newton's law has the form:
The y-component:
Then since f s   s n , we obtain
Then
42.
A rocket of mass 5 *10 4 kg is in flight. Its thrust is directed at an angle of 60
degrees above the horizontal and has a magnitude of 7 * 10 5 N . Find the
magnitude and direction of the rockets acceleration. Give the direction as an angle
above the horizontal.
Solution:
There are two forces acting on the rocket: the first one is the gravitation force and
the second one is the trust force. The net force (the sum of these two forces) will
provide an acceleration of the rocket. It is easier to find the x- and y-components
of the net force. If axis x is horizontal axis and axis y is a vertical axis then
31
Then acceleration can be found from the second Newton's law:
The magnitude of the acceleration is
The direction is characterized by the angle:
43.
The speed of a bobsled is increasing because it has an acceleration of 3 m / s 2 . At
a given instant in time, the forces resisting the motion, including kinetic friction
and air resistance, total 500 N. The mass of the bobsled and its riders is 300 kg.
(a) What is the magnitude of the force propelling the bobsled forward?
(b) What is the magnitude of the net force that acts on the bobsled?
Solution:
The net force in the sum of two forces – friction (and air resistance) force and the
force propelling the bobsled forward. There are also normal force and
gravitational force, but they cancel each other. Then
The net force provides an acceleration of the bobsled. Then
Then we can find the force F forward :
44.
A soccer ball of diameter 35 cm rolls without slipping at a linear speed of 2 m/s.
Though how many revolutions has the soccer ball turned as it moves a linear
distance of 20 m?
Solution:
When the ball makes one turn it travels the distance
32
Then if the ball travels L=20 m = 2000 cm then it makes
45.
Starting from rest, a skier slides 200 m down a 35 degrees slope. How much
longer does the run take if the coefficient of kinetic friction is 0.3 instead of 0?
Solution:
The acceleration of the skier on a slope is
With zero friction we have:
With friction 0.3 we have:
To find the traveled time we need to use the equation:
Then
and
46.
The gravitational force that the sun exerts on the moon is perpendicular to the
force that the earth exerts on the moon.
The masses are: Mass of Sun 2 *10 30 kg , Mass of Earth: 6 *10 24 kg , Mass of
Moon: 7 *10 22 kg . The distance between the sun and moon is 1 * 1011 m , and
between the moon and the earth is 4 * 10 8 m .
Determine the magnitude of the net gravitational force on the moon.
33
Solution:
The magnitude of the gravitation force that the sun exerts on the moon is
The magnitude of the gravitation force that the earth exerts on the moon is
Since the directions of the forces are orthogonal then the magnitude of the net
force is
47.
A bicycle is moving at 10 m/s. What is the angular speed of its tires if their radius
is 30 cm?
Solution:
The linear speed is related to the angular speed by the equation:
where
,
48.
. Then
A tension of 6000 Newtons is experienced by the elevator cable of an elevator
moving upwards with an acceleration of 2 m / s 2 . What is the mass of the
elevator?
Solution:
From the second Newton's law we have
Then
49.
Is it easier to move a heavy box that is sitting on the ground by (a) pulling the box
from a rope that makes an angle 300 with the surface or (b) by pushing the box
with a force that makes the same angle (but pointing downwards) with the
surface?
34
Solution:
In both cases the horizontal components of the tension force in the rope will be
the same, but the vertical components will have opposite direction.
Then the normal force in the case (a) will be less than the normal force in case (b).
Then the friction force (which is proportional to the normal force) will be less in
case (a) then in case (b).
Then it is easier to pull the box (case (a)) then to push it (case (b)).
50.
Consider a wet roadway banked, where there is a coefficient of static friction of
0.40 and a coefficient of kinematic friction of 0.2 between the tyres and the
roadway. The radius of the curve is R = 80m.
(a) If the banking angle is 300, what is the maximum speed the automobile can
have before sliding up the banking?
(b) What is the minimum speed the automobile can have before sliding down the
banking?
Solution:
(a) The direction of the forces and acceleration are shown in the figure. We write
down the second Newton's law in terms of components:
X (horizontal axis)- components:
y (vertical axis)- components:
We also have the relation between f s and n:
35
Then we have:
and
Then from the first equation (x-component):
Then
(b) In this case we have opposite direction of the friction force.
Then x and y components of the second Newton's law become:
We have the same relation between f s and n:
Then
and
Then from the first equation (x-component):
36
Then
51.
A circus clown weighs 900 N. The coefficient of static friction between the
clown's feet and the ground is 0.4. He pulls vertically downward on a rope that
passes around three pulleys and is tied around his feet. What is the minimum
pulling force that the clown must exert to yank his feet out from under himself?
Solution:
There are the following forces acting on the clown:
Tension force (applied to the hand) – direction upward
Gravitational force – direction downward,
Normal force – direction upward;
Friction force – direction horizontal;
Tension force (applied to clown's legs) – direction horizontal.
Then the minimum tension force corresponds to the condition that the friction
force is equal to 0.4*normal force.
In terms of x (horizontal) and y (vertical) components we have:
Then
52.
An athlete, who has a mass of 100 kg, can throw a 500 g ball with a speed of 10
m/s. The distance through which his hand moves as he accelerates the ball
forward from the rest until he releases it is 1.0 m. What constant force must the
athlete exert on the ball to throw it with this speed?
Solution:
If we know the acceleration then we can find the force. Therefore the first step
should be to find the acceleration of the ball. We need to use the following
equation of motion with constant acceleration
37
The initial velocity is 0, the final velocity is 10 m/s and the distance travelled is 1
m. Then we can find acceleration
Then the force is
where the mass of the ball is 500 g = 0.5 kg.
53.
A 10.0 kg block is towed up an incline at 300 with respect to the horizontal. The
rope is parallel to the incline and has a tension of 100 N. Assume that the block
starts from rest at the bottom of the hill, and neglect friction. How fast is the block
going after moving 40 m up the hill?
Solution:
To find the speed of the block we need to find the acceleration of the block. The
speed is related to acceleration and traveled distance by the following equation
where h = 40 m. The acceleration is given by the equation
Then
54.
A 0.01 kg object is moving in a plane. The x and y coordinates of the object are
given by x(t )  2t 3  t 2 and y(t )  4t 3  2t . Find the net force acting on the
object at t = 2 s.
Solution:
To find the net force we need to find acceleration of the object. The acceleration
is the second derivative of coordinate of the object. Then the x and y components
of acceleration are given by the following expressions
Then at t=2 s we have
Then the magnitude of acceleration is
38
Then the net force is
55.
An automobile weighing 3200 lbs. is on a road which rises 10 ft. for each 100 feet
of road. What force tends to move the car down the hill?
Solution:
All variables in the problem should be expressed in the same units (the same
system of units). It is better to use the SI units. In the SI system of units, the mass
should be measured in kg and the distance is in meters. Then:
The forces acting on the car are the following: gravitation force, normal force,
friction force, and the force pulling the force up along the hill. The friction force
is the rolling friction force, which is small. The normal force does not have a
component along the hill (the normal force is orthogonal to the incline). Then
there is only one force pulling the car down the hill. This is the gravitational
force, shown in the picture below. The magnitude of the gravitation force is
Then from the picture of the incline we can find the component of the
gravitational force along the inline. This component can be expressed in terms of
the angle of incline,  :
To find the angle of incline we need to use the right triangle, shown in the picture
above:
39
Then
56.
A 20 meters long rope attached to the top and the bottom of a flag pole is pulled 2
meters away from the pole by a 100 Newton force acting at right angles to the
pole at its mid point. What is the tension on the segments of the rope on each side
of the 100 Newton force?
Solution:
In problem we need to use the equilibrium condition for point A: the vector sum
of all forces is zero. There are three forces acting on point A: an external 100 N
force and two tension forces . The directions of the tension forces are along the
ropes. These forces are shown in the figure below.
The y component of the external force is zero and the y-components of the tension
forces cancel each other. Then the condition that the x-component of the sum of
all these forces is zero takes the form:
Then
40
The angle  can be found from the right triangle (with lengths of 10 m and 2 m),
shown in the figure:
Then
56. A crane cable that is capable of withstanding 22,000 N is attached by a hook to a
2,000 kg block that is resting on the ground. The cable initially starts lifting the
block at the maximum acceleration that the cable can withstand for 4 sec. It then
continues to raise the block at constant velocity for further 2 sec. At this time the
block slips off the hook at the end of the cable.
Calculate:
(1) the tension in the cable when the block is moving at constant velocity;
(2) the maximum acceleration that the cable can withstand;
(3) the maximum height that the block reaches above ground.
Solution:
(1) When the block is moving at constant velocity the acceleration of the block is
0. Then from the second Newton's law we get that the net force on the block is 0:
There are two forces acting on the block: gravitation force, pointing upward, and
the tension in the cable. The forces have opposite directions, then the magnitude
of the forces should be the same (the vector sum of these two forces is zero):
Therefore the tension in the cable is 19600 N.
(2) The maximum acceleration should be found from the condition that the
tension in the cable has its maximum value (22000 N). There are two forces
acting on the block: gravitation force and the tension in the cable. Then the
second Newton's law takes the form:
Then we rewrite this vector equation in terms of y-components (see figure below):
Then
41
(3) In the part we need to use the kinematics equations. First we have motion with
constant acceleration (we know its value from part (2)). The equations, which
describe this motion, are the following:
The initial velocity is zero: v0  0 , the initial height is zero: y0  0 . Then we
have
The block travels
. Then at this moment
Then the block moves with constant velocity. This velocity is v2  4.8 m / s . The
block travels for 2 seconds. The corresponding distance is
Then the final height of the block is
42
57.
A body of mass 10g is set to rotate in a circular path by means of a string 200 cm
long. If it makes 3 complete revolutions in 2s, find the tension of the string.
Solution:
The first step is to convert all variables into SI units:
mass should be measures in kg: m = 10 g = 0.01 kg;
length should be measured in meters: L = 200 cm = 2 m.
Now we can solve the problem. The body moves in a circle with constant speed.
This is the motion with acceleration – centripetal acceleration. The acceleration is
pointing towards the center of the circle. The magnitude of acceleration is
Here is the speed of the body (the speed is constant) and L = 2 m is the radius of
the circle, which is equal to the length of the string. Therefore, to find the
magnitude of the centripetal acceleration we need to know the speed of the body.
We know that the body makes 3 complete revolutions in 2 seconds. During one
revolution the body travels a distance of 2L , then during 3 revolutions it travels
a distance of 6L in t = 2 s. Then the speed of the body is
Now we can find the magnitude of centripetal acceleration:
The centripetal acceleration is provided by the tension in the string. Then from the
second Newton's law we obtain the tension in the string:
58.
A particle moves in a circle of radius 1 m. Its linear speed is given by v  4t ,
where t is in second and v in meter/second. Find the radial and tangential
acceleration at t  2s .
Solution:
The tangential acceleration is defined as the change of the speed (magnitude of
the velocity) of the particle. Therefore the tangential acceleration is
The radial acceleration is centripetal acceleration, which changes the direction of
velocity. The centripetal acceleration is determined by the speed (not the change
of the speed) and the radius of circular orbit:
43
The velocity at t  2s is v  4t  4  2  8 m / s , then
59.
The force required to stretch Hooke's Law spring varies from 0 N to 65 N as we
stretch the spring by moving one end 6.3 cm from its unstressed position. Find the
force constant of the spring. Answer in units of N/m.
Solution:
We understand from the formulation of the problem that the spring force is 65 N
when the spring is stretched by x  6.3 cm , which in SI units is
x  6.3 cm  0.063 m .
At the same time from the Hooke's law we know that the spring force is
where k is the force constant of the spring. We substitute the known values on the
above expression and obtain
Then
60.
A 1.5 kg mass is attached to the end of a 90 cm string. The system is whirled in a
horizontal circular path. The maximum tension that the string can withstand is
400 N. What is the maximum number of revolutions per minute allowed if the
string is not to break?
Solution:
In this problem we need to use the second Newton's law and the fact that this is a
motion with acceleration – centripetal acceleration. There are two forces acting on
the mass: gravitational force and the tension force. The vector sum of these two
forces provides the acceleration of the mass.
..................................................(1)
The centripetal acceleration is pointing towards the center of circle (rotation in a
circle).
To characterize the direction of the string we introduce angle  .
44
Now we can rewrite equation (1) in terms of x and y components:
x-component:
..................................................(2)
y-component:
Now we need to use the expression for centripetal acceleration in terms of the
angular velocity and the radius of circular orbit:
.......................................................(3)
The radius of circular orbit can be expressed in terms of the length of the string
and the angle  :
We substitute this expression in equations (2) and (3) and obtain
We can divide both sides of this equation by
45
and obtain
We know the tension has the maximum value: 400 N. Then we can find the
angular velocity:
If we know the angular velocity then we can find the period of rotation:
This is the time of one revolution. To find the number of revolutions per one
minute (60 seconds) we need to divide 60 seconds by the time of one revolution:
61.
A 20 kg child and a 30 kg child sit at opposite ends of a 4 m seesaw that is
pivoted at its center. Where should another 20 kg child sit in order to balance the
seesaw?
Solution:
In this problem we need to use only one equilibrium condition: no rotation of the
seesaw (the seesaw is balanced). The second equilibrium condition (no
translational motion) needs to be used only if we need to find the normal force at
the pivot.
We can introduce any point as the point of possible rotation. We consider point O
– the center of the seesaw – as the point of possible rotation. There is no rotation
about point O, which means that the net torque about point O is zero.
First, we need to show all forces acting on the seesaw:
1. Gravitational force (more exactly it is normal force, which is equal to the
gravitational force) m1 g  20  9.8  196 N on the 20 kg child. Application point
of the force is one end of the seesaw (see figure). The distance to point O is
d1  2 m .
2. Gravitational force (more exactly it is normal force, which is equal to the
gravitational force) m2 g  30  9.8  294 N on the 30 kg child. Application point
of the force is the other end of the seesaw (see figure). The distance to point O is
d2 .
3. We place another 20 kg child on the seesaw. We characterize the position of
the child by the distance to point O (this is an unknown distance): x . The
corresponding force is the gravitational force m3 g  20  9.8  196 N on the child.
46
There is no rotation about point O. Then the net torque about point O is zero:
We substitute the known values and obtain
From this equation we can find the position of the 20 kg child:
47
CONSERVATION LAWS
62.
A 10-kg box moving at 5 m/s on a horizontal, frictionless surface runs into a light
spring of force constant 100 N/cm. Use the work-energy theorem to find the
maximum compression of the spring.
Solution:
We need to write down the energy conservation in the present problem:
The initial energy is just the kinetic energy of the box:
The final energy is the elastic energy of the spring:
Then
and
63.
A 2000 kg truck is traveling east through an intersection at 2 m/s when it is hit
simultaneously from the side and the rear. One car is a 1000 kg compact traveling
north at 5 m/s. The other car is a 1500 kg midsize traveling east at 10 m/s. The
three vehicles become entangled and slide at one body. What are their speeds and
direction just after the collision?
Solution:
This is the example of perfectly inelastic collision: after the collision all parts of
the system move as a whole. To find the final velocity we need to use the
conservation of the net momentum.
The initial net momentum has three contributions:
- momentum of 2000 kg truck;
- momentum of 1000 kg compact;
- momentum of 1500 kg midsize.
We show these vehicles and the directions of the corresponding velocities (before
the collision).
48
Then the (initial) net momentum is
In the final state all vehicles move with the same velocity. Then the final
momentum is
Then momentum conservation takes the form
49
We need to rewrite this vector equation in terms of components. We introduce
axis x - east direction, and axis y – north direction. Then the x- component of the
above equation is
Then
The y- component of the above vector equation is
Then
Now we know x and y components of the final velocity. Then the magnitude of
the final velocity is
The direction of the final velocity is characterized by the angle between the
direction of velocity and axis x (east):
64.
A cat stuck up a tree and has 500 J gravitational potential energy. It then falls.
Find the kinetic energy of the cat just before it is caught by the owner. Find the
kinetic energy of the cat after it is caught by the owner. Find the wasted energy
(after cat is caught by the owner).
Disregard air resistance.
Solution:
As the cat falls there is a transformation of potential energy into kinetic energy.
Just before the cat is caught by the owner the kinetic energy is exactly equal to
the initial potential energy = 500 J.
After the cat is caught by the owner, its velocity is 0. It means that the kinetic
energy is equal to 0 (no motion).
The wasted energy is defined as an initial energy minus the final energy. The
initial energy is 500 J, the final energy is 0. Then the wasted energy is 500 J.
50
65.
A metal surface is illuminated one by one by photons of energy 2 eV and 4 eV
respectively. The work function of the metal is 0.5 eV. What is the ratio of the
maximum speeds of electrons emitted in two cases?
Solution:
In this problem we need to use the energy conservation. The work function of the
metal can be considered as the potential energy of electron. Then the conservation
law in the present problem takes the form: the energy of the photon is equal to the
mechanical energy of electron:
The kinetic energy is
Then
and
Then the ratio of the speeds of the electrons in two cases is
66.
A small body of mass m is attached to a light thread which passes through a hole
at the centre of a smooth table. The body is set into rotation in a circle of radius r1
with a speed of v1 . The thread is then pulled down shortening the radius of the
path to r2 . What will the new linear speed
and the new angular speed 2 be?
Solution:
In the present problem we have conservation of the angular momentum. The
initial angular momentum is
The final angular momentum is
51
Since
then
From this equation we can find the final speed of the body
The angular speed can be expressed through the linear speed and the radius of
circle:
Then
67.
Ball A, with a mass of 2 kg, moves with a velocity 5 m/s. It collides with a
stationary ball B, with a mass of 4 kg. After the collision, ball A moves in a
direction 60.0 degrees to the left of its original direction, while ball B moves in a
direction 50.0 degrees to the right of ball A's original direction. Calculate the
velocities of each ball after the collision.
Solution:
In the present problem we need to use the momentum conservation law: the net
momentum of two balls before the collision is equal to the net momentum of two
balls after the collision. It is important that momentum is a vector. Then the net
momentum of two balls is a vector sum of the momentum (vector) of ball A and
the momentum (vector) of ball B.
Therefore, the initial net momentum of two balls is
A momentum of a ball is a product of a mass of a ball and its velocity. Then
The final momentum is
Then the momentum conservation law takes the form
or
.......................(1)
52
To solve this vector equation we need to introduce coordinate system (axis x and
axis y) as shown in the figure.
We follow the standard procedure of finding the components of vectors and
rewrite the vector equation (1) in terms of x and y-components:
x-component of equation (1):
y-component of equation (1):
Finally, we have system of two equations with two unknown variables
53
From the second equation we obtain
We substitute this expression in the first equation:
Then
Therefore the velocities of the balls after collision are 22.9 m/s and 12.8 m/s (the
initial kinetic energy is less than the final energy, which means that during the
collision there is an additional source of energy).
68.
A skater of mass 80 kg initially at rest speeds up to a final speed of 10.0 m/s along
a straight line and towards the East direction.
(a) Find the momentum of the skater while at rest.
(b) Find the momentum of the skater while traveling with its final speed.
(c) Find the change in momentum of the skater.
(d) Find the impulse acted on the skater.
(e) If that impulse exerted on the skater acts for 4 s, find the average force acting
on the skater.
Solution:
In this problem we need to use the definition of momentum and impulse.
Momentum of an object is a vector, which is defined as the product of a mass of
the object and its velocity:
Therefore the direction of momentum is the same as the direction of velocity and
the magnitude of momentum is equal to the product of the mass of the object and
its speed (the magnitude of velocity).
(a) If the skater is at rest then its speed is zero. Then the momentum of the skater
is 0.
(b) The final speed of the skater is10 m/s, its mass is 80 kg, then the final
momentum is
54
(c) The change in the momentum is equal to the difference between the final
momentum and initial momentum:
(d) The impulse acted on the skater is equal to the change of its momentum:
(e) At the same time he impulse acted on the skater is equal to the product of the
(average) force acted on the skater and the time
We know time: t  4 s . Then we can find the force:
69.
A toy car “1” of mass 0.30 kg moves along a frictionless surface with a velocity
of 0.20 m/s. It collides with another toy car “2”, with a mass of 0.40 kg and a
speed of 0.10 m/s in the same direction. After the collision, toy car “1” continues
to move in the same direction with a velocity of 0.15 m/s. Calculate the speed of
toy car “2” after the collision.
Solution:
We need to use the momentum conservation law: the net momentum of two toy
cars before the collision is equal to the net momentum of two cars after the
collision.
Then the net momentum of two cars is a vector sum of the momentum of car “1”
and the momentum of car “2”.
Therefore, the initial net momentum of two cars is
A momentum of a car is a product of a mass of a car and its velocity. Then
The final momentum is
Then the momentum conservation law takes the form
or
............................(1)
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To solve this vector equation we need to introduce coordinate system. Since we
have motion along a line then it is enough to introduce only one axis: axis x (as
shown in the figure).
We follow the standard procedure of finding the components of vectors and
rewrite the vector equation (1) in terms of x-components:
x-component of equation (1):
Then
70.
A constant force of 80 N acts for 8 s on a box of mass 10 kg horizontally that
initially rests on a horizontal frictionless surface.
(a) Find the change in the box's momentum.
(b) Calculate the final speed of the box after the 8 s have passed.
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Solution:
In this problem we need to use the definition of an impulse of a force and the
relation between the impulse and the chance of the momentum of the box.
If a constant force
acts on a box for a time
then the impulse of the force is
In the present problem a constant force of 80 N acts for 8 s. Then
and
. Then the impulse is
(a) The change in the box's momentum is equal to the impulse of the force acting
on the box:
(b) The initial speed of the box is zero. Then the initial momentum of the box is
zero. The change of the momentum is equal to the difference between the final
momentum and the initial momentum (which is zero). Then
Now we know the final momentum of the box. To find the final speed we just
need to divide the final momentum by the mass of the box:
71.
At an amusement park there is a roller coaster ride ("ride of Steel"). After the first
drop, riders are moving at the speed of 120 km/h, entering an underground tunnel.
Given the fact that the roller coaster was moving at a speed of 4 km/h at the top
of the hill, determine the vertical drop that these participants fell through. Neglect
friction.
Solution:
In the problem we need to use energy conservation law: since the friction is zero
then the mechanical energy, which is the sum of kinetic energy and gravitational
potential energy, is constant.
We need to use the correct SI units, which means that the speed should be
measured in m/s. Then
Now we need to write the energy conservation law. The initial mechanical energy
is the sum of initial potential energy and the initial kinetic energy:
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The final energy is
The initial energy is equal to the final energy. Then
From the equation we obtain
We know the initial velocity, the final velocity, the initial height (it is zero). Then
we can find the final height (vertical drop):
72.
A man drops a 10 kg rock from the top of a ladder of height 5 m.
What is its speed just before it hits the ground?
What is its Kinetic Energy when it reaches the ground?
Solution:
In this problem we have free fall motion. We can solve this problem either by
using the kinematics equations, which describe the motion with constant
acceleration (free fall motion) or by using the energy conservation law.
Below we solve this problem by using the energy conservation law.
We assume that there is no friction, then we have the conservation of the net
mechanical energy of the rock. The mechanical energy of the rock is a sum of the
kinetic energy and gravitational potential energy:
The kinetic energy is determined by the speed of the rock:
The gravitational potential energy depends on the height of the rock:
Then the mechanical energy is
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The conservation of the mechanical energy means that the mechanical energy in
the initial state is equal to the mechanical energy in the final state:
or
.......................................(1)
The initial velocity is zero:
the initial height is 5 m:
the final height is zero (ground):
We do not know the final velocity:
We substitute these values in the equation (1) and obtain
Now we can find the final kinetic energy:
73.
A cyclist intends to cycle up a 8 degrees hill whose vertical height is 150 m. If
each complete revolution of the pedals moves the bike 6 m along its path,
calculate the average force that must be exerted on the pedals tangent to their
circular path. Neglect work done by friction and other losses. The pedals turn in a
circle of diameter 30 cm. The total mass of the cyclist and his bike is 100 kg.
Solution:
In this problem we need to use generalized work-energy theorem: work done by
an external force is equal to the change of the net mechanical energy of the
system:
We assume that the cyclist moves with constant speed. Then the initial and the
final kinetic energies are the same. Therefore in the above expression we need to
take into account only the gravitational potential energy:
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where
is the total mass of the cyclist and bike.
Then
Now we need to define the initial and the final states of the system. We introduce
the final state as the state after one complete revolution (relative to the initial
state).
We know that after one complete revolution the cyclist moves 6 meters (as
shown in the figure). Then the change in the height of the cyclist is
Therefore the change in the gravitation potential energy of the cyclist + bike is
The work done by the force is
During one revolution the pedal travels a distance of
(circumference of a circle with diameter d). Where d = 30 cm = 0.3 cm is the
diameter of the circle. Then the work done by the force is
This work is equal to the change in gravitational potential energy:
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From this expression we can find the force:
74.
A body of mass 5 kg slides a distance of 6 m down a rough Inclined plane 30
degree. Then it moves on frictionless horizontal surface and compresses a spring.
The coefficient of kinetic friction is 0.1 and the spring constant is 300 N/m. Find
the maximum compression of the spring.
Solution:
In this problem we need to use the work-energy theorem, which determines the
relation between the work done by the friction force and the change of the net
mechanical energy.
According the work-energy theorem the work done by the friction force (which is
always negative) is equal to the difference between the final mechanical energy
and the initial mechanical energy.
................................................. (1)
The mechanical energy is the sum of kinetic energy, gravitational potential
energy, and elastic energy of the spring.
The initial velocity of the body is zero, the initial compression of the spring is
zero, the initial height of the body is
initial mechanical energy of the body is
. Then the
In the final state we have a maximum compression of the spring. It means that in
the final state the velocity of the body is zero. The final height of the body is 0
(ground level). Then the final mechanical energy of the body is
Then equation (1) takes the form
Now we need to find the work done by the friction force. The work done by the
friction force (see figure) is
where s = 6 m is the displacement of the body over incline plane. The friction
force is determined by the normal force and the coefficient of kinetic friction:
61
To find the normal force we need to write down the second Newton's law for the
motion along the incline. There are three forces acting on the body: gravitational
force, normal force, and the friction force. Then the second Newton's law takes
the form:
...........................................................(2)
The direction of the acceleration is along the incline. It means that the
acceleration has only x component (see figure). The y-component of acceleration
is zero. Then to find the normal force we need to write down only the ycomponent of equation (2):
From this equation we can find the normal force:
Now we can find the friction force
and the work done by the friction force
Then we can find the maximum (final) compression of the spring:
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75.
A uniform, solid metal disk of mass 6.0 kg and diameter 2.0 cm hangs in a
horizontal plane, supported at its center by a vertical metal wire. You find that it
requires a horizontal force of 4 N tangential to the rim of the disk to turn it by 5
degrees, thus twisting the wire. You now remove this force and release the disk
from rest.
(a) What is the torsion constant for the metal wire?
(b) Write the equation of motion for twist angle of the disk.
Solution:
(a) The torque is related to the torsion constant (k) by the equation:
We know that when we apply the force 4 N the system is in equilibrium with a
twist angle
. The torque in this case is
. Then
(b) To write equation of motion we need to find the moment of inertia of the disk.
It has the following expression:
Then
or
The tallest spot on Earth is Mt. Everest, which is 8857 m above sea level. If the
radius of the Earth to sea level is 6369 km, how much does the magnitude of g
change between sea level and the top of Mt. Everest?
Solution:
The value of the free fall acceleration is
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Then
Or
76.
The value of g at the surface of the earth is 9.78 N/kg, and on the surface of
Venus the magnitude of g is 8.6 N/kg. A cosmonaut has a mass of 60 kg on the
surface of the earth. What will her weight be on the surface of Venus?
Solution:
The weight is the product of the mass of cosmonaut and the free fall acceleration.
On the surface of Venus:
77.
A car (m=2000 kg) is parked on a road that rises 20 degrees above the horizontal.
What are the magnitudes of (1) the normal force and (2) the static frictional force
that the ground exerts on the tires?
Solution:
We need to write down the second Newton's law: the net force is 0. This is the
vector equation. In the present problem there are three forces acting on the car:
gravitational force, normal force, and the static friction force.
Then we rewrite the second Newton's law in terms of x and y-components (x axis
is parallel to the road).
The x-component of the second Newton's law has the form:
The y-component:
From these equations we can find the normal force:
and the static friction force:
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78.
A spaceship is on a straight-line path between Earth and moon. At what distance
from the Earth is net gravitational force on the spaceship zero?
Solution:
Let us introduce the mass of the Earth as M earth , the mass of the moon as M moon ,
and the distance between the Earth and the moon as R .
If the spaceship is at distance x from the Earth, then the gravitation force on the
spaceship due to Earth is
The gravitation force on the spaceship due to moon is
Since the net gravitation force should be zero, then we have the equation:
From this equation we have:
and
79.
The space shuttle is orbiting the Earth at a distance of about 300 km from its
surface. At that distance, the gravitational acceleration is almost the same as that
on the surface. How long does it take for the shuttle to complete one orbit around
the Earth? Assume that the orbit is circular.
Solution:
The gravitation force will provide the acceleration to the shuttle, which is equal
where
Then a  g and
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Then the period is
80.
When a falling object is at a distance above the Earth's surface of 4 times the
Earth's radius, what is its free-fall acceleration due to the Earth's gravitational
force exerted on it?
Solution:
The gravitation force has the following expression
Here M is the mass of the Earth, R is the radius of the Earth, h is the distance
above the Earth. Since h  4R , then
Then acceleration is
Since g  G
M
 9.8 m / s 2 , then
2
R
The drawing shows three particles far away from any other objects and located on
a straight line. The masses of these particles are: particle A = 400 kg, particle B =
500 kg, and particle C = 100 kg. Find the magnitude and direction of the net
gravitational force acting on each of the three particles (the direction to the right is
positive).
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Solution:
Particle A: The net gravitational force is the sum of two forces: due to particle B
and due to particle C. Both forces have positive direction:
Particle B: The net gravitational force is the sum of two forces: due to particle A
(this force is negative, which means that its direction is to the left) and due to
particle C (this force is positive). Then
Particle C: The net gravitational force is the sum of two forces: due to particle A
and due to particle B. Both forces are negative, then
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