# Solutions - MIT OpenCourseWare

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```Momentum and Impulse
Concept Tests
Question 1 Consider two carts, of masses m and 2m, at rest on an air track. If you push
first one cart for 3 s and then the other for the same length of time, exerting equal force
on each, the momentum of the light cart is
1.
2.
3.
4.
5.
four times
twice
equal to
one-half
one-quarter
the momentum of the heavy cart.
Answer 3. The change in momentum is equal to the impulse. For a constant force, the
impulse is equal to the product of the force with the time interval the force is applied..
Because the forces on the carts are equal, as are the times over which the forces act, the
impulses applied to each cart are equal hence the change in momenta are equal. Since
both carts start from rest, the final momenta of the two carts are equal.
Question 2 Pushing Identical Carts
Identical constant forces push two identical objects A and B continuously from a starting line to
a finish line. If A is initially at rest and B is initially moving to the right,
1. Object A has the larger change in momentum.
2. Object B has the larger change in momentum.
3. Both objects have the same change in momentum
4. Not enough information is given to decide.
Answer 1: Both objects have the same mass, are pushed the same distance, by the same constant
force, so they have the same acceleration. Since object B has an initial speed the time interval
needed to reach the finish is less than the corresponding time interval for object A which started
from rest. Therefore the change in velocity of object B is less than the corresponding change in
velocity for object A. Hence object A has a larger change in momentum.
Question 3 Pushing Non-Identical Carts
Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart for 3 s
and then the other for the same length of time, exerting equal force on each, the kinetic energy of
the light cart is
1. larger than
2. equal to
3. smaller than
the kinetic energy of the heavy car.
Answer 1. The kinetic energy of an object can be written as
K=
1 2 p2
mv =
2
2m
Because the impulse is the same for the two carts, the change in momentum is the same. Both
start from rest so they both have the same final momentum. Since the mass of the lighter cart is
smaller than the mass of the heavier cart, the kinetic energy of the light cart is larger than the
kinetic energy of the heavy cart.
Question 4: Same Momentum Different Masses
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same
momentum, and you exert the same force to stop each. How do the distances needed to stop them
compare?
1. It takes a shorter distance to stop the ping-pong ball.
2. Both take the same distance.
3. It takes a longer distance to stop the ping-pong ball.
4. Not enough information is given to decide.
Since the ping pong ball and the bowling ball have the same momentum, the kinetic energy of
the less massive ping pong ball is greater than the kinetic energy of the more massive bowling
ball since K = p 2 / 2m . You must do work on an object to change its kinetic energy. If you exert
a constant force, then the work done is the product of the force with the displacement of the point
of application of the force. Since the work done on an object is equal to t he change in kinetic
energy, the ping pong ball has a greater change in kinetic energy in order to bring it to a stop, so
the you need a longer distance to stop the ping pong ball.
Alternative Solution: Both the initial momentum and the force acting on the two objects are
equal. Therefore the initial velocity and the acceleration of the ping-pong ball isgreater than the
bowling ball by the ratio of the bowling ball mass to the ping-pong ball mass.
m p v0, p = mb v0,b = px
m p a p = mb ab = Fx
Since both the force acting on each object and the change in momentum is the same, the impulse
acting on each ball is the same. Therefore, the time interval it takes to stop each object is the
same. Since the displacement is equal to
a !t #
"
!x = % vx,0 \$ x & !t
2 (
'
The ratio
!x p mb
=
!xb m p
hence the ping-pong has the greater displacement.
Question 5
A ball is thrown against a wall; the ball bounces off and returns with speed equal to the speed it
had before striking the wall. Which of the following statements is true from before to after the
1) The kinetic energy of the ball is the same.
2) The momentum of the ball is the same.
3) Both the kinetic energy and the momentum of the ball are the same.
4) Neither the kinetic energy nor the momentum of the ball are the same.
5) The collision is inelastic.
6) Two of the above.
Solutions:
1) True – speed is given as unchanged.
2) Not true – momentum is a vector, and the direction of the ball’s direction is reversed, and the
momentum cannot be the same after bouncing off the wall. The magnitude of the momentum is
the same, but that’s not what this question asks.
3,4) Can’t be, since 1) is true and 2) is not.
5) Not true - if the speed is the same, the collision is elastic.
6) Not true, since only 1) is true.
Question 6 The figure below depicts the paths of two colliding steel balls, A and B.
Which of the arrows 1-5 best represents the impulse applied to ball B by ball A during the
Solution:
1; Ball B has changed its momentum in the upward direction in the figure, and as far as the
figure can show, there is no change in its horizontal (rightward) velocity.
Question 7
A force of magnitude F is applied to a dumbbell first as in (a) and then as in (b). In which case
does the dumbbell acquire the greater center-of-mass acceleration? Explain your reasoning.
Solution
It doesn’t matter. For practical purposes, it might be hard to maintain a force in a constant
direction for case (b), as the dumbbell would tend to rotate (we’ll deal with this when we get to
rotational motion, of course), but as long as the forces are applied in the same direction, the
acceleration of the center of mass will be the same. You might anticipate (or you might have
seen in some other physics class) that in case (b) the upper end of the dumbbell, where the force
is applied, will initially move twice as fast as the center of mass. Still, the center of mass will
accelerate the same in both cases.
Question 8
Skater A of mass 75 kg and skater B of mass 50 kg are initially at rest some distance apart. Each
skater holds tightly onto a rope of negligible mass. Skater A pulls on the rope with constant force
so that the skaters approach each other and meet. The ice is completely frictionless. Which
statements below are true and which are false? Explain your reasoning.
1. Only skater B moves relative to the ice.
2. The magnitude of the acceleration of skater A is less than the magnitude of the
acceleration of skater B.
3. Just before they meet, the speed of skater A is less than the speed of B.
4. While the skaters are moving, their momentum vectors have equal magnitudes.
Solution
Before giving the answers, consider that saying that “Skater A only pulls on the rope” is perhaps
misleading (not intentionally, of course). The force that skater A exerts on the rope and the force
that the rope exerts on skater A are an action-reaction pair, as are the forces that skater B and the
rope exert on each other. Thus, to hold onto the rope, skater B must be pulling on the rope as
well; as the problem states, skater B “holds tightly” to the rope. If we assume a massless rope,
as given in the problem statement, then the magnitude of the net force on each skater is the same.
Thus, statement (1) is false. Statements (2) and (3) are true, and of course are sort of the same
statement, in slightly different forms, given that both skaters are initially at rest. Since the mass
of the rope is neglected, the magnitude of the net force is the same, as mentioned above, and as
there is no net force on the skaters/rope system, the momenta of the skaters must be equal in
magnitude (but opposite in direction).
MIT OpenCourseWare
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8.01SC Physics I: Classical Mechanics