Download Chapter 2: Kinematics in One Dimension Example

Exam Review
Tuesday, September 17, 2013
10:00 PM
Chapter 2: Kinematics in One Dimension
Example: A juggler throws a ball straight up with an initial speed of 10 m/s.
With what speed would she need to throw a second ball, half a second later,
starting from the same position as the first ball, so that the second ball hits the
first ball at the top of the first ball's trajectory?
Solution:
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Example: A car starts from rest at a stop sign. It accelerates at
2.0 m/s2 for 6.0 s, then coasts for 2.0 s, then slows down at a rate of 1.5 m/s2
until the next stop sign is reached. Determine the distance between the stop
signs.
Solution:
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Chapter 3: Kinematics in Two Dimensions
Example: A car travelling at 30 m/s runs out of gas while travelling up a 5.0
degree slope. How far will it coast before starting to roll back down?
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Example: On the Apollo 14 mission to the Moon, astronaut Alan Shepard hit a golf ball
with a golf club improvised from a tool. The free-fall acceleration on the surface of the
Moon is 1/6th that of the Earth. Suppose he hit the ball with an initial speed of 25 m/s at
an angle of 30 degrees above the horizontal.
a. Determine how long the ball was in flight.
b. Determine how far the ball travelled (horizontally).
c. Ignoring air resistance, how much farther would the ball travel on the Moon
than on the Earth?
c. Repeat the calculation for Earth data, and you'll find that both the time of flight
and the horizontal distance are 6 times greater on the Moon.
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Example: A spring-loaded gun, fired vertically, shoots a marble 6.0 m straight
up. Determine the marble's range if it is fired horizontally from 1.5 m above
the ground.
Solution: First determine the initial velocity of the marble.
Next, determine the time of flight for the second motion.
Therefore, the range is
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Example: A car moves around a circular road that has radius 110 m.
a. Determine the car's acceleration if the car moves through the curve at a constant
speed of 40 m/s.
b. At what speed would the car's acceleration be double the value calculated in Part
a?
Solution:
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Example: A child slides along frictionless ice at a speed of 4 m/s relative to the
ice at an angle of 31 degrees north of east. The child throws a puck along the
ice at a speed of 2 m/s relative to the child at an angle of 26 degrees south of
east. Determine the velocity of the puck relative to the ice.
Solution: The easiest way to add the velocity vectors of the child and the puck
is to separate each of them into components. Here are the details:
If you prefer magnitude and direction:
Thus, the velocity of the puck relative to the ice is (5.23, 1.18) m/s, which is
equivalent to 5.36 m/s in the direction E13◦N.
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Example: Car A moves at a constant speed of 80 km/h East relative to the
ground, and Car B moves at a constant speed of 70 km/h north relative to the
ground. Determine the velocity of Car B relative to Car A.
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Solution:
If you prefer magnitude and direction:
Thus, the velocity of Car B relative to Car A is (−80, 70) km/h, which is
equivalent to 106 km/h in the direction E41◦N.
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Example: A person stands 10 m away from the base of a wall that is 8 m high.
She can throw comfortably at an angle of 60 degrees above the horizontal.
Determine the minimum initial speed she must give a tennis ball so that it
will clear the wall.
Solution: Draw a diagram!
Consider the points on the ball's path labelled A and B; they are the key
points of the path. Because we know the coordinates of both points, it
makes sense to use the displacement equations, as follows:
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makes sense to use the displacement equations, as follows:
We have two equations for the two unknown quantities, so we have
enough to solve the problem. One way to solve the equations is to solve
equation (2) for t and substitute the resulting expression into equation
(1), which can then be solved for the required initial speed:
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Chapter 4: Forces and Newton's Laws of Motion
Example: A 500 kg piano is being lowered into position by a crane while two
people steady it with ropes pulling to the sides. Bob's rope pulls to the left, 15
degrees below the horizontal, with 500 N of tension. Ellen's rope pulls to the
right, 25 degrees below the horizontal.
a. Determine the tension in Ellen's rope if the piano descends vertically at a
constant speed.
b. Determine the tension in the main cable supporting the piano.
Solution:
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Example: Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s.
The distance through which his hand moves as he accelerates the rock forward from
rest until he releases it is 1.0 m.
a. Determine the constant force Bob exerts on the rock.
b. If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?
Solution: Bob clearly has a "rifle-arm."
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Example: A person with compromised pinch strength in his fingers can only
exert a normal force of 6.0 N to either side of a pinch-held object, such as a
book. Determine the heaviest book he can hold if the coefficient of static
friction between his fingers and the surface of the book is 0.80.
Solution:
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Example: A wood block is sliding up a wood ramp. If the ramp is very steep, the
block will reverse direction at its highest point and slide back down. If the ramp
is shallow, the block will stop when it reaches its highest point. Determine the
smallest ramp angle, measured from the horizontal, for which the block will
slide back down. (Note that the coefficient of static friction is 0.5 and the
coefficient of kinetic friction is 0.2.)
Solution: Once the block reaches its highest point, the forces acting on it along
the slope are a component of its weight and the frictional force. If the slope is
not large enough, then static friction will be able to balance the component of
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not large enough, then static friction will be able to balance the component of
weight along the slope. When the static friction force is maximum,
For larger angles, the block will slide back down, as the static friction force will not be
large enough to balance the component of the weight directed down the slope.
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Here are the free-body diagrams for a problem on a previous test:
The full solution to the problem is posted online; examine it carefully if you
wish.
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wish.
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Example: In the drawing, the rope and the pulleys are massless, and there is
no friction. Find (a) the tension in the rope and (b) the acceleration of the
10.0-kg block. (Hint: The larger mass moves twice as far as the smaller
mass.)
Solution: Draw free-body diagrams
for each block!
Let x1 represent the displacement of
the more massive block, and let x2
represent the displacement of the
less massive block. The hint means
that (using the chosen positive
directions):
Because each block begins from rest,
it follows that
Applying Newton's second law of motion
to each block (with the help of the freebody diagrams), we obtain
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Chapter 5: Dynamics of Uniform Circular Motion
Example: A car drives over the top of a hill that has a radius of 50 m. Determine
the car's maximum speed so that it does not fly off the road at the top of the hill.
Solution:
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Example: A 100 g ball on a 60-cm-long string is swung in a vertical circle whose
centre is 200 cm above the floor. The string suddenly breaks when it is parallel to
the ground and the ball is moving upward. The ball reaches a height 600 cm
above the floor. Determine the tension in the string an instant before it broke.
Solution:
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Example: A sensitive gravimeter at a mountain observatory finds that the freefall acceleration is 0.0075 m/s2 less than at sea level. Determine the
observatory's altitude.
Solution:
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Example: Suppose we could shrink the Earth without changing its mass. At what
fraction of its current radius would the free-fall acceleration at the surface be
three times its current value?
Solution:
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Example: Determine the speed and altitude of a geostationary
satellite orbiting Mars. Mars rotates on its axis once every 24.8 h, has a mass of
6.42 × 1023 kg, and has a radius of 3370 km.
Solution:
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Chapter 6: Work and Energy
Example: A swing is made from a rope that can support a maximum tension
of 800 N without breaking. Initially, the swing hangs vertically. The swing is
then pulled back to an angle of 60 degrees with respect to the vertical and
released from rest. Determine the mass of the heaviest person that can
ride the swing without breaking the rope.
Solution: Draw a free-body diagram! But at which point of the swing? I'm
not sure, so I'll draw a free-body diagram for a random point of the swing.
Then I'll write down Newton's second law of motion for the radial
component and the tangential component of the motion:
We are asked to determine something related
to the maximum tension in the rope. You can
see the relation between the maximum tension
and the maximum mass allowable most easily
by solving equation (1) for the tension:
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Can you see that the tension is at its maximum when the person is at the lowest
point of the motion? That is the point when the speed is greatest, and it's also the
point where the cosine of the angle is the greatest, because the cosine of 0
degrees is 1. Thus,
Now if we only had a way of knowing the maximum speed (i.e., the speed of the
mass when it reaches the lowest point of its motion), then we could make further
progress. The other problem is that we don't know r, the length of the rope. I'm not
sure what to do about this latter problem, but for the former problem, I would
definitely try energy methods. For instance, upon reading the problem again, I
notice that we haven't used the fact that the initial angle is 60 degrees. This calls for
another diagram, and an application of the principle of conservation of mechanical
energy. (Recall that the tension force in this case does no work on the person,
because the tension force is always radial, which is perpendicular to the motion.)
Now substitute this relation into the key
equation from above,
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and we obtain:
The swing is clearly not save, even for children, much less for more massive
adults.
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Chapter 7: Impulse and Momentum
Example: A firecracker in a coconut blows the coconut into three pieces. Two
pieces of equal mass fly off south and west, perpendicular to each other, at 20
m/s. The third piece has twice the mass as the other two. Determine the speed
and direction of the third piece.
Solution:
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Example: A 10 g bullet is fired into a 10 kg wood block that is at rest on a
wooden table. The block, with the bullet embedded, slides 5.0 cm across the
table. Determine the speed of the bullet. (The coefficient of friction is 0.20.)
Solution:
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Example: A 1500 kg weather rocket accelerates upward at 10.0 m/s2. It explodes
2.00 s after liftoff and breaks into two fragments, one twice as massive as the other.
Photos reveal that the lighter fragment traveled straight up and reached a
maximum height of 530 m. What were the speed and direction of the heavier
fragment just after the explosion?
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Example: The figure shows a collision between three balls of clay. The three hit
simultaneously and stick together. Determine the speed and direction of the
resulting blob of clay.
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Example: A 20 g ball is fired horizontally toward a 100 g ball that is hanging
motionless from a 1.0-m-long string. The balls undergo a head-on, elastic
collision, after which the 100 g ball swings out to a maximum angle of 50 degrees.
Determine the initial speed of the 20 g ball.
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Example: Two blocks, A and B, slide on a frictionless surface. Block A has
an initial velocity of 8 m/s at an angle of 20 degrees south of east, and
Block B has an initial velocity v at an angle of 30 degrees north of east. The
blocks collide; after the collision, Block A has a velocity of 5 m/s at an
angle of 50 degrees north of east and Block B has a velocity of 7 m/s at an
angle of 40 degrees south of east. The mass of Block B is 1.3 kg. Determine
the mass m of Block A and the speed v of Block B before the collision.
Solution: Use the principle of conservation of momentum.
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We have two equations in two unknowns, which is enough to solve the problem.
One way to do this is to solve each equation for the unknown speed, and then
equate the two expressions:
Setting the expressions in equations (3) and (4) equal to each other, and solving for
m, we obtain
Substituting the value for the mass of Block A into equation (3), we obtain the speed of
Block B before the collision:
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Thus, the mass of Block A is 1.1 kg, and the speed of Block B before the collision is
2.0 m/s.
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Chapter 8: Rotational Kinematics, and Chapter 9: Rotational Dynamics
Example: The 2.5 kg object shown in the figure has a moment of inertia about
the rotation axis of 0.085 kg  m2. The rotation axis is horizontal. When released,
what will be the magnitude of the object's initial angular acceleration?
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Example: A computer disk is 8.0 cm in diameter. A reference dot on the edge of
the disk is initially located at an angle of 45 degrees. The disk accelerates steadily
for 0.50 s, reaching 2000 rpm, then coasts at a steady angular velocity for another
0.50 s.
a. Determine the tangential acceleration of the reference dot after 0.25 s.
b. Determine the centripetal acceleration of the reference dot after 0.25 s.
c. Determine the angular position of the reference dot after 1.0 s.
d. Determine the speed of the reference dot after 1.0 s.
Solution:
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Example: The 20-cm-diameter disk in the figure can rotate on an axle through its
centre. Determine the net torque about the axle.
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Example: The ropes in the figure are each wrapped around a cylinder, and the
cylinders are fastened together. The smaller cylinder has a diameter of 10 cm
and a mass of 5.0 kg; the larger cylinder has a diameter of 20 cm and a mass of
20 kg. Determine the angular acceleration of the cylinders assuming they turn
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20 kg. Determine the angular acceleration of the cylinders assuming they turn
on a frictionless axle.
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Example: A bicycle is rolling down a circular portion of a path, as shown in
the figure. This portion of the path has radius 9.00 m. The angular
displacement of the bicycle is 0.96 rad. Each bicycle wheel has radius 0.400
m. Determine the angle through which each bicycle wheel turns.
Solution: First determine the distance travelled by the bicycle:
Now note that the distance travelled
by a spot on one of the tires is the
same as the distance just calculated.
This allows us to calculate the angle
that each tire turns through, as
follows:
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Example: Consider two identical coins lying flat on a table. One coin is
fixed in place and the second coin is touching the first coin. The
movable coin is rotated in such a way that it always touches the fixed
coin, and rolls along it without slipping. When the movable coin is
moved all the way around the fixed coin so that it returns to its
starting position, through what angle has the moving coin turned?
Solution: The moving coin makes two complete rotations, so it moves
through an angle of 720 degrees. Study the following diagrams:
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Example: A tennis ball, starting from rest, rolls (without slipping) down
the hill in the drawing. At the end of the hill the ball becomes airborne,
leaving at an angle of 35 degrees with respect to the horizontal. Treat the
ball as a thin-walled spherical shell, and determine the range x.
Discussion: Our goal is to determine the translational velocity of the ball
at point 2. If we can do this, then the rest of the problem is just a
projectile motion problem, of the type that we have solved back in
Chapter 3.
Solution: The moment of
inertia of the ball is
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Solution: The moment of
inertia of the ball is
Assuming that the ball's
mechanical energy is
conserved between points
1 and 2 in the diagram,
Because of the no-slipping condition,
Thus,
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Now that we know the translational speed at position 2, and we know
the projection angle, we can solve the projectile motion problem to
determine the range.
First determine the time of flight, and then use it to determine the range.
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Chapter 10: Simple Harmonic Motion
Example: How far must you stretch a spring with stiffness constant 1000 N/m to
store 200 J of energy?
Solution:
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Solution:
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Example: A 10 kg runaway grocery cart runs into a spring with stiffness constant
250 N/m and compresses it by 60 cm. What was the speed of the cart just
before it hit the spring?
Solution:
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Some conceptual questions
When a ball is thrown vertically upward, the force acting on the ball in a vertically
upward direction gradually decreases as the ball's speed decreases.
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When a ball is thrown vertically upward it gradually slows down, momentarily
stops, and then falls down again. When it momentarily stops, the net force acting
on the ball is zero.
A cargo plane flies West at 900 km/h. When the plane is directly over the Brock
tower, it drops a package of physics textbooks. The package lands to the West of
the tower.
When a heavy truck collides with a light car, the force that the truck exerts on
the car is greater than the force that the car exerts on the truck.
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An elevator is lifted up at a constant speed by a steel cable. The force that the
cable exerts on the elevator cabin is greater than the force that gravity exerts on
the elevator cabin.
A passenger in a car moving very fast around a circular curve feels that he or
she is "thrown" towards the outside of the curve because of a force that pushes
objects away from the centre of the circular curve.
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If we assume no air resistance, then for a projectile motion the net force in the
horizontal direction is constant but the net force in the vertical direction is not
constant.
A horse is hitched to a cart. The driver says, "Giddyup!", but the horse doesn't move. He
argues (yes, he's a talking horse) that there is no point in exerting any effort, because no
motion is possible, because of Newton's third law. "After all," says the horse, "when I exert
a force on the cart, by Newton's third law the cart exerts an equal and opposite force on
me. The net force is therefore zero, and nothing moves." Explain.
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A 250-pound linebacker collides with a 150-pound wide receiver. The receiver is thrown to the
ground violently. Obviously the force that the linebacker exerts on the wide receiver is greater
than the force that the wide receiver exerts on the linebacker.
When you fire a rifle, you should hold the butt of the rifle tightly against your shoulder.
Why?
(a) Explain how a propeller (boat or airplane) works.
(b) How do rockets work?
If the acceleration vector of a moving object is non-zero and NOT in the same
direction as the velocity vector, then the object's speed is NOT increasing.
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direction as the velocity vector, then the object's speed is NOT increasing.
A ball is thrown straight up. At the peak of its motion, the ball stops momentarily.
Because the ball is stopped momentarily, it experiences no net force for that
instant.
In a collision between a truck and a car, if the truck is twice as massive as the
car then the truck exerts a force on the car that is twice as large as the force that
the car exerts on the truck.
Rockets work in space in the same way that garden hoses recoil when water surges
through them; that is, the hot gases ejected from the rocket engine nozzles at very
high speed cause the rocket to move in the opposite direction.
For an object moving along a curved path, the net force acts along a tangent to
the curve, because this force is needed to push the object along its path.
If it's di cult to turn a bolt, then it's helpful to switch to a longer wrench, because
the longer wrench is more massive and therefore has greater force.
The angular speed of a spinning skater increases when she brings her arms in; this
illustrates the principle of conservation of angular momentum.
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In space stations that are above the Earth's atmosphere, astronauts are weightless
because they are beyond the reach of Earth's gravity.
According to the special theory of relativity, everything is relative, not absolute
as was previously thought in Newtonian mechanics.
Energy is always conserved, but momentum is not always conserved.
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