Download b) Electromagnetic Force

C. Forces
1. What is force?
Force is a push or pull on an object resulting from the object's interaction with
another object. For example, if you push a car a force is exerted on the car. If
the applied force of pushing is greater than the resistance force of friction the car
will move. Forces are vectors and therefore have both magnitude and direction.
Forces acting upon an object are usually shown using a free-body diagram such as the example
below
2. Four Fundamental Forces (from greatest to weakest)
a) Strong Nuclear Force
The force that binds particles in the nucleus together. Without this force matter could not
form.
b) Electromagnetic Force
The force that gives materials their strength and ability to bend, squeeze, stretch, or
shatter. This force comes from the particles electric charge. Particles have electrostatic
forces whether at motion or at rest, but produce both electrostatic and magnetic forces
when they are in motion.
c) Weak Nuclear Force
A form of electromagnetic force that involves the radioactive decay of some nuclei
causing for example beta decay.
d) Gravitational Force
The attractive force between all objects.
The following chart indicates the relative strength of each fundamental force compared to the
strong nuclear force and the relative distances that each force will interact.
Fundamental Forces
It is currently believed that all four of the fundamental forces are, in fact, part of a single
unifying force. While no such force has been discovered the theory behind it is referred to as
the Unified Field Theory (UFT)
Four Fundamental Forces.url
3. Newton’s Three Laws of Motion
a) 1st Law
An object with no net force (balanced) acting upon it
remains at rest or moves with constant velocity in a
straight direction. The 1st law was developed from
Galileo's concept of inertia. Inertia according to Galileo
is an objects tendency to proceed in a straight direction
and not change its motion unless other forces are
applied.
b) 2nd Law
The acceleration of a body is directly proportional to the net
force (unbalanced) acting upon the body and inversely
proportional to the objects mass. The 2nd law can be stated
using the following formula





 Fnet Where F is force (N), m is mass (kg) and a is acceleration
2
Fnet  ma  a 
to describe force is called the Newton, where
m (m/s ). The unit
1N = 1kgm/s2
If an object has a net force acting on it, it will move in the direction of the net force. Forces to
the right (horizontal) and upwards (vertical) are positive, and those to the left or down are
negative. All forces that are applied to the right are added to produce one large force, and the
same for the other forces. The sum of all these forces will indicate the magnitude and direction
of the net force (recall addition of vectors). If the magnitude is "zero", there is no movement.
Example #1
A person pushes a desk with a mass of 45kg. If the person accelerates the desk at 1.3m/s2,
determine the force applied on the object.
Solution

Fnet


m = 45kg
a = 1.3m/s2
Fnet = ?

 ma  45kg  1.3m / s 2  59kgm / s 2  59N


Example #2
A crate whose mass is 360kg rests on the bed of a truck that is moving at a speed of 120km/h.
The driver applies the brakes and slows down to 62km/h in17s. What net force acts on the
crate during this time?
Solution
  
v f  v i  at
 
 v f  v i 17.22m / s  33.33m / s  16.11m / s
a


 0.9477m / s 2
t
17
s
17
s


Fnet  ma  360kg   0.9477m / s2  341.1kgm / s2  340N


c) 3rd Law
When one object exerts a force on a second object, the second
object exerts a force on the first that is equal in magnitude, but
opposite in direction.
For example, if you push against a wall with a force of 100N then
the wall will push back at you with the same force but in the
opposite direction.
4. Mass and Weight
a) What is mass?
Mass is the amount of matter an object contains and is usually denoted in kg. Mass
does not change regardless where you are.
b) What is weight?
Weight is the force of gravity acting on a particular mass and is denoted in Newton’s.
Weight changes depending on the force of gravity at a particular location.



FW  W  mg
For example, if the mass of an object is 60.0kg, determine its weight.
Solution

W  mg  60.0kg   9.81m / s 2  588kgm / s 2  588N
NOTE: Even though weight is vector quantity it is stated in terms of its magnitude
and therefore is always given as a positive value.

Assignment #1

Mass and weight.url
Lab #1 – Force
5. Friction
Friction is the force that opposes movement of an object. For example - when a tire rolls
across the ground, the force of friction will stop the tire from rolling.
a) Two types of friction
i) Static Friction – is the frictional force acting between two surfaces which are
attempting to move, but are not moving. It is the force that opposes the start of motion.
ii) Kinetic Friction – is the frictional force acting between two surfaces which are in
motion against each other.
Static friction and kinetic friction are both types of friction: they act when two surfaces are
moving or attempting to move against each other, and resist the motion. The main difference
between static and kinetic friction is that static friction acts while the surfaces are at rest while
kinetic friction acts when there is relative motion between the surfaces.
Static friction is the frictional force acting between two surfaces which are attempting to move,
but are not moving. Think of having a block of wood on a table. If you attach a string to it and
pull it with a very small force, it would not move. Since you are applying a force in the
“forward” direction, according to Newton’s first law there must be another “backward” force to
maintain the object at rest. This is the frictional force. It acts along the two surfaces, in the
direction opposite to the direction of attempted motion.
Now imagine increasing the pulling force. For a while, the block would still not move. In order
for the forces to remain balanced, the force of friction must be also increasing as you increase
the force of the pull. Eventually, at some point, your pulling force will be able to overcome
friction and the object will begin to move. When the object just begins to move, the force of
static friction has reached its maximum value.
Kinetic friction is the frictional force acting between two surfaces which are in motion against
each other. Going back to our example of dragging a wooden block across a table, once the
block is moving it will experience kinetic friction.
When using free-body diagrams, the opposite force to friction is usually referred to as the
applied force. Weight is the downward force acting upon the object and the opposite force to
weight is the normal force.
The force of friction depends on the force pushing the surface together FN (normal force) and
the characteristics of the surfaces. These characteristics are summarized by the term
coefficient of friction and is denoted by the symbol µ. The coefficient of friction changes
depending on the type of surface. The higher the value for µ the greater the resistance to
motion and therefore its frictional properties. The equation for the force of friction is


Ff  μFN


Where Ff is the force of friction, µ is the coefficient of friction and FN is the normal force.
Example
A 3.0kg toy is being pulled by a force of 24N. If it starts from rest, how far will it travel in the
first 5.0s?
Solution



 Fnet
Fnet  ma  a 
m
 24N
24kg  m / s 2
a

 8.0m / s 2
3.0kg
3.0kg
 
1
d  v i t  at 2
2

1
2
d  0m / s 5.0s   8.0m / s 2 5.0s   100m  1.0 x102 m
2

Assignment #2, #3 and #4
Lab #2 – Pulleys

7. Pulleys
Pulleys are simple machines that consist of a rope that slides around a
disk, called a block. Their main function is to change the direction of the
tension force in a rope. The pulley systems consist of massless and
frictionless pulleys, and ropes that are massless and that don’t stretch.
These somewhat unrealistic parameters mean that:
i) The rope slides without any resistance over the pulley, so that the
pulley changes the direction of the tension force without changing its
magnitude.
ii) You don’t have to factor in the mass of the pulley or rope when
calculating the effect of a force exerted on an object attached to a
pulley system.
a) The Purpose of Pulleys
We use pulleys to lift objects because they reduce the amount of
force we need to exert. For example, say that you are applying
force F to the mass in the figure at the right. How does F compare
T
to the force you would have to exert in the absence of a pulley?
F
To lift mass m at a constant velocity without a pulley, you would
have to apply a force equal to the mass’s weight, or a force of mg
upward. Using a pulley, the mass must still be lifted with a force
mg
of mg upward, but this force is distributed between the tension of
the rope attached to the ceiling, T, and the tension of the rope gripped in your hand, F.
Because there are two ropes pulling the block, and hence the mass, upward, there are
two equal upward forces, F and T. We know that the sum of these forces is equal to the
gravitational force pulling the mass down, so F + T = 2F = mg or F = mg/2. Therefore,
you need to pull with only one half the force you would have to use to lift mass m if there
were no pulley.
b) Standard Pulley Problem
The figure to the right represents a pulley system where
masses m1 and m2 are connected by a rope over a
massless and frictionless pulley. Note that m1 > m2 and both
masses are at the same height above the ground. The
system is initially held at rest, and is then released. From
experience, we know that the heavy mass, m1, will fall, lifting
the smaller mass, m2. Because the masses are connected,
we know that the velocity of mass m1 is equal in magnitude
to the velocity of mass m2, but opposite in direction.
Likewise, the acceleration of mass m1 is equal in magnitude
to the acceleration of mass m2, but opposite in direction.

Note that the tension force F , on each of the blocks is of the same magnitude. In any
T
non-stretching rope the tension, as well as the velocity and acceleration, is the same at
every point.
c) What is the acceleration of mass m1?
Because the acceleration of the rope is of the same magnitude at every point in the
rope, the acceleration of the two masses will also be of equal magnitude. If we label the
acceleration of mass m2 as a, then the acceleration of mass m1 is –a (opposite
direction).
For mass m1:



FT  m1g  m1a

and F  m g  m a
T
1
1
For mass m2:



FT  m2g  m2a

and F  m g  m a
T
2
2
Therefore




m2g  m2a  m1g  m1a




m1a  m2a  m1g  m2g



am1  m2   m1g  m2g


 m1g  m2g
a
m1  m2 

 g m1  m2 
a
m1  m2 
Examples
Assignment #5
8. Momentum
Let’s look at momentum by considering two vehicles of equal
mass. A heavy truck moving rapidly has a large momentum. It
takes a large or prolonged force to get the truck up to this
speed, and it takes a large or prolonged force to bring it to a
stop afterwards. If the truck were lighter, or moving more
slowly, then it would have less momentum. Therefore the
velocity and mass of the vehicle determine the force needed to
change its motion. The product of the mass
 and velocity is
called momentum. It is represented by p and is determined using the formula


p  mv where


p is momentum, m is mass, v is velocity
The unit for momentum is kgm/s or Ns and is a vector quantity.
a) Impulse
When a force is applied over a period of
time the momentum of the object changes.
The product of the force and the time
interval over which it acts ( Ft ) is called
the impulse and is found using the


formula Ft  mv where


m is mass, v is the change in velocity, t is the change in time, F is the applied force.
The unit for impulse is the Ns.
In general, if the mass remains constant then the momentum must change
 whenthere




is a change in velocity. Therefore p  mv and since Ft  mv then Ft  p which
is referred to as the impulse-momentum theorem.
b) Conservation of Momentum
The law of conservation of momentum states that the total
momentum of a closed system does not change. This means
that when two objects collide the total momentum of the objects
before the collision is the same as the total momentum of the
objects after the collision. Total momentum means the
momentum of object 1 plus the momentum of object 2. Object
1 might not have the same momentum before and after the
collision but the amount of momentum object 1 gains from the
collision is the same as the amount of momentum object 2
loses so the total momentum remains the same whether the
collision is elastic or inelastic. The total momentum is also the
same before and after an explosion.
i) What does Closed System mean?
"Closed system" is used to mean that only the forces of the two objects colliding
affects the motion of the objects. See Newton's third law of motion. If another force
(called an external force) affects the motion of the objects then the momentum is
not conserved, it will change.
Examples
Understanding Car Crashes It's Basic Physics.url
Assignment #6
C. Forces
Assignment #1
1. When a shot-putter exerts a net force of 140N on a shot, the shot has an acceleration of 19m/s2. What is the
mass of the shot? 7.4kg
2. Together a motorbike and rider have a mass of 275kg. The motorbike is slowed down with an acceleration of 4.50m/s2. What is the net force on the motorbike? Describe the direction of this force and the meaning of the
negative sign. -1.24 x 103N
3. A car, mass 1225kg, traveling at 105km/h, slows to a stop in 53m. What is the size and direction of the force that
acted on the car? What provided the force? -9.8 x 103N opposite to the cars motion Friction
4. Imagine a spider with mass 7.0 x 10-5kg moving downward on its thread. The thread exerts a force that results in
a net upward force on the spider of 1.2 x 10-4N. What is the acceleration of the spider? 1.7m/s2
5. What is the weight of each of the following objects?
a) 0.113kg hockey puck. 1.11N
b) 108kg football player. 1.06 x 103N
c) 870kg automobile. 8.50 x 103N
6. Find the mass of each of these weights.
a) 98.0N. 9.99kg
b) 80.0N. 8.15kg
c) 0.98N. 0.10kg
7. A 20N stone rests on a table. What is the force the table exerts on the stone and in what direction is this force
being applied? 20N upward
8. An astronaut with a mass of 75kg travels to Mars. What is the astronauts weight on
a) Earth? 7.4 x 102N
b) Mars where g = -3.8m/s2? 2.9 x 102N
9. Suppose Joe, who weighs 625N, stands on a bathroom scale calibrated in Newton’s.
a) What force would the scale exert on Joe and in what direction? 625N upward
b) If Joe now holds a 50.0N cat in his arms, what force would the scale exert on him? 675N
c) After Joe puts down the cat, his father comes up behind him and lifts upward on his elbows with a 72.0N
force. What force does the scale now exert on Joe? 553N upward
10. An object with a mass of 15kg rests on a frictionless horizontal plane and is acted upon by a horizontal force of
30.0N. Determine:
a) What is the acceleration? 2.0m/s2
b) How far will it move in 10.0s? 1.0 x 102m
c) What will be its velocity after 10.0s? 2.0 x 10m/s
11. A car with a mass of 1.0 x 103kg is moving in a straight line at a constant velocity of 30.0m/s. It is brought to rest
in 25s. What constant force is acting to stop the car? -1200N
12. A 3.0kg toy is pulled by a force of 24N. If it starts from rest, how far will it travel in the first 5.0s? 1.0 x 102m
13. An 8.0g bullet travelling at 4.0 x 102m/s passes through a heavy wood block in 4.0 x 10-4s, emerging with a
velocity of 1.0 x 102m/s. Ignore any motion of the wood.
a) With what average force did the wood oppose the motion of the bullet? -6.0 x 103N
b) How thick is the block of wood? 0.10m
14. A 0.22 caliber rifle shoots a bullet of mass 1.8g with a muzzle velocity of 5.0 x 102m/s. If the barrel is 25cm long,
what is the average force exerted on the bullet while it is in the barrel? 9.0 x 102N
15. A sled of mass 6.0kg is moving along a smooth, horizontal ice surface with a velocity of vi. A force of 36N is
applied to the sled in its direction of motion, increasing its velocity to 17.6m/s while moving it 10.0m. What was the
sled's original velocity? 14m/s
16. A force of 8.0N gives a mass, m1, an acceleration of 2.0m/s2 and a mass, m2, an acceleration of 4.0m/s2. What
acceleration would the force give the two masses if they were fastened together? 1.3m/s2
Assignment #2
1. The driver of a 1500kg car puts on the brakes on a concrete road. Calculate the force of friction.
a) On a dry road. (µ = 0.70) 1.0 x 104N
b) On a wet road. (µ = 0.30) 4400N
2. A 70.0kg box is slid along the floor by a 400.0N force. The coefficient of friction between the box and the floor is
0.50 when the box is sliding. Determine the box's rate of acceleration. 0.81m/s2
3. A tractor moving at a constant velocity is ploughing a field pulling a 100.0kg plough with a force of 880N. What is
the coefficient of friction? 0.90
4. A 600.0kg car is moving on a level road at 3.0 x 10m/s.
a) How large of a retarding force is required to stop it in a distance of 70.0m? -3900N
b) What is the coefficient of friction between the tires and the road? 0.66
5. A 100.0N block rests on the floor and the coefficient of sliding friction between the block and the floor is 0.250. A
horizontal force of 40.0N acts on the block accelerating it for 3.00s. Determine the velocity of the block after
3.00s. 4.41m/s
6. A force of 260.0N pushes on a 25.0kg block starting from rest, then achieves a velocity of 2.00m/s in a time of
4.00s. Determine the coefficient of sliding friction between the box and the floor. 1.01
7. It takes a 50.0N horizontal force to pull a 20.0kg object along the ground at a constant velocity. What is the
coefficient of friction? 0.255
8. A cart of mass 2.0kg is pulled across a level desk by a horizontal force of 4.0N. If the coefficient of friction is 0.12:
a) What is the acceleration of the cart? 0.83m/s2
b) If the cart started from rest what would be its velocity after the force has acted for 2.00s? 1.7m/s
9. A 52N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36N is exerted. What is
the coefficient of sliding friction between the sidewalk and the metal runners of the sled? 0.69
10. Suppose the sled now runs on packed snow. The coefficient of friction is now only 0.12. If a person weighing
650N sits on the sled, what force is needed to slide the sled across the snow at constant speed? 84N
11. The coefficient of sliding friction between rubber tires and wet pavement is 0.50. The brakes are applied to a
750kg car traveling 30.0m/s, and the car skids to a stop.
a) What is the size and direction of the force of friction that the road exerts on the car? -3.7 x 103N
b) What would be the size and direction of the acceleration of the car? -4.9m/s2
c) How far would the car travel before stopping? 92m
12. If the tires of the car in #11 did not skid, the coefficient of friction would have been 0.70. Would the force of friction
have been larger, smaller, or the same? Would the car have come to a stop in a shorter, the same, or a longer
distance? larger, shorter
Assignment #3
1. Determine the acceleration of a 13140kg V2 rocket fired vertically, if the upward thrust of the engines is 2.6 x
105N. 9.9m/s2 up
2. An elevator has an upward acceleration of 1.0m/s2. With what force will the elevator floor push upward on an 82kg
passenger? 890N up
3. After re-entry, an astronaut of mass 75kg is rescued at sea on the end of a rope attacked to a helicopter. When
the helicopter accelerates upward at 0.50m/s2, what is the tension (upward force that the rope must withstand)?
770N up
4. Matilda, who weighs 5.80 x 102N, plans to elope by sliding down an improvised rope made of nylon stockings.
The maximum tension the stockings can withstand without tearing is 400.0N.
a) Can she slide down with a uniform velocity? No
b) What is the minimum acceleration with which she can slide down the rope? 3.05m/s2 dn
5. A stone of mass 1.0kg falls to the bottom of a lake with an acceleration of 5.0m/s2. What is the upward buoyancy
force of the water? 4.8N up
6. An 88kg aviator during free fall acquires a speed of 46m/s and then opens his parachute. Ten seconds after it
opens his speed is 6.6m/s. Determine the upward force supplied by the parachute. 1200N up
7. A force of 90.0N is exerted when lifting a stone weighing 7.35N. Calculate the final velocity after 5.0s if the stone
was initially at rest. 550m/s up
8. A rocket weighing 7840N on earth is fired. The force of propulsion is 10440N. Determine the distance travelled
after 10.0s. 163m up
9. A 5.0kg object is lifted upward a distance of 20.0m acquiring a final speed of 10.0m/s. Determine the force
required to do this. 62N up
Assignment #4
1. A dirt bike climbs at a steady speed up a hill which is inclined at 60.0° from the vertical. Sketch a free-body
diagram related to the situation.
2. Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with a mass of 0.440kg, is on
the table. Becca lifts her end of the table before Ryan does and as a result, the table makes an angle of 15.0°,
with the horizontal. Determine the components of the cup’s weight that are parallel and perpendicular to the plane
of the table. F|| = – 1.12N F ⊥ = – 4.17N
3. Fernando, who has a mass of 43.0kg, slides down the banister at his grandparent’s house. If the banister makes
an angle of 35.0° with the horizontal, determine the normal force between Fernando and the banister. 346N
4. A 562N crate is resting on a plane inclined 30.0° above the horizontal. Determine the components of the crate’s
weight that are parallel and perpendicular to the plane. F|| = – 281N F ⊥= – 487N
5. Determine the speed of the crate from question #4 after 4.00s. -19.6m/s
6. Lynda has a mass of 45.0kg. She slides down a slide sloped at 27.0°. The coefficient of kinetic friction is 0.230.
Determine her rate of descent 1.00s after starting from rest. -2.44m/s
7. Stacie who has a mass of 45.0kg starts down a slide that is inclined at an angle of 45° with the horizontal. If the
coefficient of kinetic friction between Stacie’s and the slide is 0.25, determine her acceleration. 5.2m/s2
8. A rope pulls a 63kg water skier up a 14.0° incline with a tension of 512N. The coefficient of kinetic friction between
the skier and the ramp is 0.27. Determine the magnitude and direction of the skier’s acceleration. 3.2m/s2 14°
9. A 215N box is placed on an incline that makes a 35.0° angle with the horizontal. Determine the component of the
weight parallel to the incline. F|| = -176N F ⊥= -123N
Assignment #5
1. Two spheres of masses are tied together by a light string looped over a frictionless pulley. If they
are allowed to hang freely, determine:
a) The acceleration if M = 3.00kg and m = 1.50kg. 3.27m/s2 lt
b) The acceleration if M = 7.00kg and m = 2.00kg. 5.44m/s2 lt
2. A 3.00kg mass is attached to a 5.00kg mass by a strong string that passes over a frictionless pulley. When the
masses are allowed to hang freely, what will be:
a) The acceleration of the masses? 2.45m/s2 rt
b) The tension on the string? 36.8N
3. Calculate the accelerations in the systems below.
a) 25kg
b) 15kg
µ = 0.28
µ = 0.37
15kg
3.7m/s2 rt
c)
11kg
6.0kg
10.0kg
µ = 0.20
9.0kg
12kg
0.97m/s2 rt
0.39m/s2 rt
4. Two blocks are connected by a sting over a frictionless and massless pulley such
that one is resting on an inline plane and the other is hanging over the top edge of
the plane as shown in the diagram to the right. The mass of m1 and m2 are 16.0kg
and 8.0kg respectively. If the coefficient of kinetic friction is 0.23, the angle of
inclination is 37.0° and the blocks are released from rest, determine:
a) The acceleration of the block.
b) The tension in the string connecting the blocks.
Assignment #6
1. A 0.144kg baseball is pitched horizontally at 38m/s. After it is hit by a bat, it moves horizontally at –381m/s.
a) Determine the impulse the bat delivered to the ball. -11kgm/s
b) If the bat and ball were in contact 8.0 x 10-4s, determine the average force the bat exerted on the ball.
-1.4 x 104N
c) Determine the average acceleration of the ball during its contact with the bat. -9.7 x 104m/s2
2. A compact car, mass 725kg, is moving at 100.0km/h.
a) Determine the momentum of the car. 2.02 x 104kgm/s
b) At what velocity is the momentum of a larger car with a mass 2175kg equal to that of the smaller car?
33.4km/h
3. A snowmobile has a mass of 2.50 x 102kg. A constant force is exerted on it for 60.0s. The snowmobile’s initial
velocity is 6.00m/s and its final velocity 28m/s. Determine:
a) The change in momentum. 5.5 x 103kgm/s
b) The magnitude of the force exerted on it? 92N
4. The brakes exert 6.40 x 102N force on a car weighing 15680N and moving at 20.0m/s. The car finally stops.
Determine:
a) The car's mass? 1.60 x 103kg
b) The car's initial momentum. 3.20 x 104kgm/s
c) The change in the cars momentum. -3.20 x 104kgm/s
d) The length of time the braking force must be applied to bring the car to rest? 50.0s
5. A 0.105kg hockey puck moving at 48m/s is caught by a 75kg goalie at rest. With what speed does the goalie slide
on the ice? 0.067m/s
6. A 35.0g bullet strikes a 5.0kg stationary block of wood and embeds itself in the block. The block and bullet fly off
together at 8.6m/s. Determine the original velocity of the bullet. 1.2 x 103m/s
7. A 35.0g bullet moving at 475m/s strikes a 2.5kg wooden block. The bullet passes through the block leaving at
175m/s. If the block was at rest when it was hit, determine the speed of the bullet when it exits. 2.8m/s
8. A 0.50kg ball traveling at 6.0m/s collides head-on with a 1.00kg ball moving in the opposite direction at a velocity
of –12.0m/s. The 0.50kg ball moves away at –14m/s after the collision. Determine the velocity of the second ball.
-2.0m/s
9. A 15g bullet is shot into a 5085g wooden block standing on a frictionless surface. The block, with the bullet in it,
acquire a velocity of 1.0m/s. Calculate the velocity of the bullet before it strikes the block. 340m/s
10. A hockey puck, mass 0.115kg, moving at 35.0m/s, strikes a rubber octopus thrown on the ice by a fan. The
octopus has a mass of 0.265kg. If the puck and octopus slide off together, determine their final velocity. 10.6m/s
11. A 92kg fullback, running at 5.0m/s, attempts to dive across the goal line for a touchdown. Just as he reaches the
goal line, he is met head-on in midair by two 75kg linebackers, one moving at 2.0m/s and the other at 4.0m/s. If
they all become entangled as one mass, with what velocity do they travel? Does the fullback score? 0.041m/s
TOUCHDOWN
12. A 5.00g bullet is fired with a velocity of 100.0m/s toward a 10.00kg stationary solid block resting on a frictionless
surface. Determine:
a) The change the in momentum of the bullet if it is embedded in the block. -0.500kgm/s
b) The change in momentum of the bullet if it ricochets in the opposite direction with a speed of 99m/s.
-0.995kgm/s
13. A 150g baseball pitched at a speed of 41.6m/s is hit straight back to the pitcher at a speed of 61.5m/s. The bat
was in contact with the ball for 4.70 x 10-3s. Determine the average force exerted by the bat on the ball. 3290N
14. A 2500kg unmanned space probe is moving in a straight line at a constant speed of 300.0m/s. A rocket engine on
the space probe executes a burn in which a thrust of 3000.0N acts for 65.0s. Determine the change in momentum
of the craft. 1.95 x 105kgm/s
15. Two parts of a spacecraft are separated by detonating the explosive bolts that hold them together. The masses of
the parts are 1200kg and 1800kg. If an impulse of 300.0Ns is applied determine the relative speed of recession of
the two parts. 0.41.7m/s
Review
1. A person weighing 490N stands on a scale in an elevator.
a) What does the scale read when the elevator is at rest?
b) The elevator ascends accelerating the person upward at 2m/s2. What does the scale read now?
c) The elevator reaches a constant speed. What is the reading on the scale as the elevator rises uniformly?
d) The elevator slows down as it reaches the proper floor. Do the scale readings increase or decrease?
e) The elevator descends. Does the scale reading increase or decrease?
f) What does the scale read if the elevator descends at a constant speed?
g) If the cable snaps and the elevator falls freely, what would the scale read?
2. If twice the force gives twice the acceleration, why does a 10kg stone not fall twice as fast as a 5kg stone?
3. A boy jumps down from a platform with a pound weight on his upturned hand. What will happen to the force on his
hand:
a) While it is in the air?
b) When he lands on the ground?
4. If a suitcase falls from a luggage rack of a fast moving train, will it fall on the passenger in the seat in front, in the
seat below, or in the seat behind?
5. UFO's were perplexing since they seemed to be able to make angular turns without changing speed. What
physical law did they seem to violate?
6. If you release the air from an inflated toy balloon as you suddenly let go of it, what happens? Explain.
7. If two men were dropped together by a helicopter on to the middle of a completely frictionless surface, how could
they get off? Could one man get off he were dropped alone?
8. A force of 20.0N gives a rock a horizontal acceleration along a frictionless surface of 4.0m/s2. Calculate the mass
of the rock. 5.0kg
9. Determine the weight of an 8.0kg object. 78N
10. A 1500kg car moving along a level highway experiences a road friction of 900.0N and an air resistance of
3000.0N. If the wheels exert a force of 5000.0N, determine the rate of acceleration. 0.73m/s2
11. A 10.0kg object is pushed horizontally so that its velocity changes from 10.0m/s to 25.0m/s over a distance of
30.0m. If the force of friction was 1000.0N, determine the initial horizontal force. 1090N
12. Determine the coefficient of friction for a 20.0kg object being moved across a surface at constant velocity if the
applied friction is 120N. 0.61
13. A rocket weighs 4900N. What upward force would produce an acceleration of 4.0m/s2? 6900N
14. Determine the vertical acceleration produced when a 0.50kg ball is thrown straight upwards with a force of 75N.
140m/s2
15. Calculate the vertical force required to change the velocity of an object moving upward from 10.0m/s to 30.0m/s
over a distance of 25m. (the mass of the object is 0.30kg) 4.8N
16. A force is exerted horizontally on a 2.0kg ball causing it to travel a distance of 25m in the 2.0s that the force is
being applied. Determine the value of the force if the coefficient of friction between the ball and the surface is
0.050. 26N
17. Determine the tension in the cable of a 500.0kg elevator which is accelerating towards the ground at 5.0m/s2.
2400N
18. Determine the acceleration produced in each of the situations below.
a)
1.96m/s2 lt
b)
4.3m/s2 lt
3.0kg
µ = 0.10
15.0kg
10.0kg
9.4kg
2.5kg
19. At Sea World, a 900.0kg polar bear slides down a wet slide inclined at an angle of 25.0° to the horizontal. The
coefficient of friction between the bear and the slide is 0.0500. What frictional force impedes the bear's motion
down the slide?
20. Rather than taking the stairs, Martin gets from the second floor of his house to the first floor by sliding down the
banister that is inclined at an angle of 30.0° to the horizontal.
a) If Martin has a mass of 45kg and the coefficient of friction between Martin and the banister is 0.20, what is
the force of friction impeding Martin's motion down the banister?
b) If the banister is made steeper, will this have any effect on the force of friction? If so, what?
21. A chunk of rock mass 50.0kg slides down the side of a volcano that slopes up at an angle of 30.0° to the
horizontal. If the rock accelerates at a rate of 3.0m/s2, what is the coefficient of friction between the rock and the
side of the volcano?
22. A 1250kg slippery hippo slides down a mud covered hill inclined at an angle of 18° to the horizontal.
a) If the coefficient of friction between the hippo and the mud is 0.0900, what force of friction impedes the
hippo's motion down the hill?
b) If the hill were steeper, how would this affect the coefficient of friction?
23. A 0.160kg puck is travelling at 5.0m/s. A slapshot produces a collision that lasts for 0.0020s and gives the puck a
final velocity of 50.0m/s.
a) Determine the impulse imparted by the hockey stick. 7.2Ns
b) Determine the average force applied by the stick to the puck. 3.6 x 103N
24. Calculate the momentum of:
a) A 2.5kg rabbit travelling with a velocity of 2.0m/s. 5.0kgm/s
b) A 5.0kg groundhog travelling with a velocity of 1.0m/s. 5.0kgm/s
25. A hockey player of mass 97kg skating with a velocity of 9.2m/s collides head-on with a defence player of mass
105kg travelling with a velocity of 6.5m/s. An instant after the impact, the two skate together in the same direction.
Calculate the final velocity of the two hockey players. 1.0m/s
26. Consider a collision that involves two objects of masses 4.5kg and 6.2kg. The larger mass is initially at rest and
the smaller mass has an initial velocity of 16m/s. The final velocity of the larger object is 10.0m/s. Calculate the
final velocity of the smaller object after the collision. 2.2m/s