End of chapter exercises | Momentum and impulse

Momentum

Textbook Exercise 2.7

[SC 2003/11]A projectile is fired vertically upwards from the ground. At the highest point of its motion, the projectile explodes and separates into two pieces of equal mass. If one of the pieces is projected vertically upwards after the explosion, the second piece will...

drop to the ground at zero initial speed.
be projected downwards at the same initial speed as the first piece.
be projected upwards at the same initial speed as the first piece.
be projected downwards at twice the initial speed as the first piece.

be projected downwards at the same initial speed as the first piece.

[IEB 2004/11 HG1] A ball hits a wall horizontally with a speed of $\text{15}$ $\text{m·s$^{-1}$}$. It rebounds horizontally with a speed of $\text{8}$ $\text{m·s$^{-1}$}$. Which of the following statements about the system of the ball and the wall is true?

The total linear momentum of the system is not conserved during this collision.
The law of conservation of energy does not apply to this system.
The change in momentum of the wall is equal to the change in momentum of the ball.
Energy is transferred from the ball to the wall.

The change in momentum of the wall is equal to the change in momentum of the ball.

[IEB 2001/11 HG1] A block of mass M collides with a stationary block of mass 2M. The two blocks move off together with a velocity of $\vec{v}$. What is the velocity of the block of mass M immediately before it collides with the block of mass 2M?

$\vec{v}$
2$\vec{v}$
3$\vec{v}$
4$\vec{v}$

\begin{align*} M\vec{v}_{1i} + 2M\vec{v}_{2i} & = (M+2M)\vec{v} \\ M\vec{v}_{1i} & = 3M\vec{v} \\ \vec{v}_{1i} & = 3\vec{v} \end{align*}

3$\vec{v}$

[IEB 2003/11 HG1] A cricket ball and a tennis ball move horizontally towards you with the same momentum. A cricket ball has greater mass than a tennis ball. You apply the same force in stopping each ball.

How does the time taken to stop each ball compare?

It will take longer to stop the cricket ball.
It will take longer to stop the tennis ball.
It will take the same time to stop each of the balls.
One cannot say how long without knowing the kind of collision the ball has when stopping.

Since their momenta are the same, and the stopping force applied to them is the same, it will take the same time to stop each of the balls.

[IEB 2004/11 HG1] Two identical billiard balls collide head-on with each other. The first ball hits the second ball with a speed of V, and the second ball hits the first ball with a speed of 2V. After the collision, the first ball moves off in the opposite direction with a speed of 2V. Which expression correctly gives the speed of the second ball after the collision?

[SC 2002/11 HG1] Which one of the following physical quantities is the same as the rate of change of momentum?

resultant force
work
power
impulse

resultant force

[IEB 2005/11 HG] Cart X moves along a smooth track with momentum p. A resultant force F applied to the cart stops it in time t. Another cart Y has only half the mass of X, but it has the same momentum p.

In what time will cart Y be brought to rest when the same resultant force F acts on it?

$\frac{1}{2}t$
$t$
$2t$
$4t$

$t$

[SC 2002/03 HG1] A ball with mass m strikes a wall perpendicularly with a speed, v. If it rebounds in the opposite direction with the same speed, v, the magnitude of the change in momentum will be ...

$2mv$
$mv$
$\frac{1}{2}mv$
$0mv$

$2mv$

Show that impulse and momentum have the same units.

The units of momentum are $\text{kg⋅m⋅s$^{-1}$}$:

Impulse can be defined as force over total time: $\text{N⋅s}$. $\text{1}\text{ N} = \text{1}\text{ kg⋅m⋅s$^{-2}$}$. Therefore the units for impulse are: $\text{kg⋅m⋅s$^{-1}$}$

This is the same as the units for momentum.

the change in the momentum of the golf ball.

\begin{align*} \Delta p & = F_{\text{net}}\Delta t \\ & = (\text{3} \times \text{10}^{\text{3}})(\text{5} \times \text{10}^{-\text{4}}) \\ & = \text{1,5}\text{ kg⋅m⋅s$^{-1}$} \end{align*}

the velocity of the golf ball as it leaves the club.

\begin{align*} \Delta p & = mv \\ \text{1,5} & = \text{0,06}v \\ v & = \text{25}\text{ m⋅s$^{-1}$} \end{align*}

What does the area under this graph represent?

Impulse

Calculate the speed at which the ball leaves the hockey stick.

\begin{align*} \text{Impulse } & = F \Delta t \\ & = \Delta p = m \Delta v \end{align*}

The impulse is the area under the graph:

\begin{align*} \text{Impulse } & = (\text{0,5})(150)(\text{0,5}) \\ & = \text{37,5}\text{ N} \end{align*}

The speed is:

\begin{align*} \Delta v & = \frac{\text{37,5}}{\text{0,150}} \\ & = \text{250}\text{ m⋅s$^{-1}$} \end{align*}

The same player hits a practice ball of the same mass, but which is made from a softer material. The hit is such that the ball moves off with the same speed as before. How will the area, the height and the base of the triangle that forms the graph, compare with that of the original ball?

The area will remain the same because the final velocity and the mass are the same. The duration of the contact between the bat and the ball will be longer as the ball is soft, so the base will be wider. In order for the area to be the same, the height must be lower. Therefore, the player can hit the softer ball with less force to impart the same velocity on the ball.

The fronts of modern cars are deliberately designed in such a way that in case of a head-on collision, the front would crumple. Why is it desirable that the front of the car should crumple?

If the front crumples then the force of the collision is reduced. The energy of the collision would go into making the front of the car crumple and so the passengers in the car would feel less force.

[SC 2002/11 HG1] In a railway shunting yard, a locomotive of mass $\text{4 000}$ $\text{kg}$, travelling due east at a velocity of $\text{1,5}$ $\text{m·s$^{-1}$}$, collides with a stationary goods wagon of mass $\text{3 000}$ $\text{kg}$ in an attempt to couple with it. The coupling fails and instead the goods wagon moves due east with a velocity of $\text{2,8}$ $\text{m·s$^{-1}$}$.

Calculate the magnitude and direction of the velocity of the locomotive immediately after collision.

\begin{align*} m_{1}v_{i1} + m_{2}v_{i2} & = m_{1}v_{f1} + m_{2}v_{f2} \\ (\text{4 000})(\text{1,5}) & = (\text{3 000})(\text{2,8}) + (\text{4 000})v_{f2} \\ v_{f2} & = -\text{0,6}\text{ m⋅s$^{-1}$} \\ & = \text{0,6}\text{ m⋅s$^{-1}$} \text{ west} \end{align*}

Name and state in words the law you used to answer the previous question

The principle of conservation of linear momentum. The total linear momentum of an isolated system is constant.

[SC 2005/11 SG1] A combination of trolley A (fitted with a spring) of mass $\text{1}$ $\text{kg}$, and trolley B of mass $\text{2}$ $\text{kg}$, moves to the right at $\text{3}$ $\text{m·s$^{-1}$}$ along a frictionless, horizontal surface. The spring is kept compressed between the two trolleys.

While the combination of the two trolleys is moving at $\text{3}$ $\text{m·s$^{-1}$}$, the spring is released and when it has expanded completely, the $\text{2}$ $\text{kg}$ trolley is then moving to the right at $\text{4,7}$ $\text{m·s$^{-1}$}$ as shown below.

State, in words, the principle of conservation of linear momentum.

The total linear momentum of an isolated system is constant.

Calculate the magnitude and direction of the velocity of the $\text{1}$ $\text{kg}$ trolley immediately after the spring has expanded completely.

\begin{align*} (m_{1}+m_{2})\vec{v}_i & = m_{1}\vec{v}_{f1} + m_{2}\vec{v}_{f2} \\ (\text{2}+\text{1})(\text{3}) & = (\text{2})(\text{4,7}) + (\text{1})\vec{v}_{f2} \\ \vec{v}_{f2} & = -\text{0,4}\text{ m·s$^{-1}$} \end{align*}

$\vec{v}_{f2} = \text{0,4}\text{ m·s$^{-1}$}$ to the left

$\text{0,4}$ $\text{m·s$^{-1}$}$ to the left

[IEB 2002/11 HG1] A ball bounces back from the ground. Which of the following statements is true of this event?

The magnitude of the change in momentum of the ball is equal to the magnitude of the change in momentum of the Earth.
The magnitude of the impulse experienced by the ball is greater than the magnitude of the impulse experienced by the Earth.
The speed of the ball before the collision will always be equal to the speed of the ball after the collision.
Only the ball experiences a change in momentum during this event.

The magnitude of the change in momentum of the ball is equal to the magnitude of the change in momentum of the Earth.

[SC 2002/11 SG] A boy is standing in a small stationary boat. He throws his schoolbag, mass $\text{2}$ $\text{kg}$, horizontally towards the jetty with a velocity of $\text{5}$ $\text{m·s$^{-1}$}$. The combined mass of the boy and the boat is $\text{50}$ $\text{kg}$.

Calculate the magnitude of the horizontal momentum of the bag immediately after the boy has thrown it.

\begin{align*} p & = mv \\ & = (2)(5) \\ & = \text{10}\text{ kg·m·s$^{-1}$} \end{align*}

Calculate the velocity (magnitude and direction) of the boat-and-boy immediately after the bag is thrown.

\begin{align*} 0 & = m_{1}\vec{v}_{1f} + m_2\vec{v}_{2f} \\ -\text{10}& = (\text{50})\vec{v}_{2f}\\ \vec{v}_{2f} & = \frac{-\text{10}}{\text{50}}\\ & = -\text{0,2}\text{ m·s$^{-1}$} \end{align*}

$\text{0,2}$ $\text{m·s$^{-1}$}$ in the opposite direction to the jetty

Chapter summary

Table of Contents

3.1 Introduction