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Physics In Action":" Impulse

2.6 Physics in action: Impulse (ESCJM)

A very important application of impulse is improving safety and reducing injuries. In many cases, an object needs to be brought to rest from a certain initial velocity. This means there is a certain specified change in momentum. If the time during which the momentum changes can be increased then the force that must be applied will be less and so it will cause less damage. This is the principle behind arrestor beds for trucks, airbags, and bending your knees when you jump off a chair and land on the ground.

Air-bags in motor vehicles (ESCJN)

Air bags are used in motor vehicles because they are able to reduce the effect of the force experienced by a person during an accident. Air bags extend the time required to stop the momentum of the driver and passenger. During a collision, the motion of the driver and passenger carries them towards the windshield. If they are stopped by a collision with the windshield, it would result in a large force exerted over a short time in order to bring them to a stop. If instead of hitting the windshield, the driver and passenger hit an air bag, then the time of the impact is increased. Increasing the time of the impact results in a decrease in the force.

\begin{align*} F& = \frac{\Delta p}{\Delta t} \end{align*}

Therefore if t is increased, for a constant change in momentum, the force on the body is reduced.

Padding as protection during sports (ESCJP)

The same principle explains why wicket keepers in cricket use padded gloves or why there are padded mats in gymnastics. In cricket, when the wicket keeper catches the ball, the padding is slightly compressible, thus reducing the effect of the force on the wicket keepers hands. Similarly, if a gymnast falls, the padding compresses and reduces the effect of the force on the gymnast's body.

Arrestor beds for trucks (ESCJQ)

An arrestor bed is a patch of ground that is softer than the road. Trucks use these when they have to make an emergency stop. When a trucks reaches an arrestor bed the time interval over which the momentum is changed is increased. This decreases the force and causes the truck to slow down.

Follow-through in sports (ESCJR)

In sports where rackets and bats are used, like tennis, cricket, squash, badminton and baseball, the hitter is often encouraged to follow-through when striking the ball. High speed films of the collisions between bats/rackets and balls have shown that following through increases the time over which the collision between the racket/bat and ball occurs. This increase in the time of the collision causes an increase in the velocity change of the ball. This means that a hitter can cause the ball to leave the racket/bat faster by following through. In these sports, returning the ball with a higher velocity often increases the chances of success.

Crumple zones in cars (ESCJS)

Another safety application of trying to reduce the force experienced is in crumple zones in cars. When two cars have a collision, two things can happen:

1. the cars bounce off each other, or

2. the cars crumple together.

Which situation is more dangerous for the occupants of the cars? When cars bounce off each other, or rebound, there is a larger change in momentum and therefore a larger impulse. A larger impulse means that a greater force is experienced by the occupants of the cars. When cars crumple together, there is a smaller change in momentum and therefore a smaller impulse. The smaller impulse means that the occupants of the cars experience a smaller force. Car manufacturers use this idea and design crumple zones into cars, such that the car has a greater chance of crumpling than rebounding in a collision. Also, when the car crumples, the change in the car's momentum happens over a longer time. Both these effects result in a smaller force on the occupants of the car, thereby increasing their chances of survival.

Egg Throw

This activity demonstrates the effect of impulse and how it is used to improve safety. Have two learners hold up a bed sheet or large piece of fabric. Then toss an egg at the sheet. The egg should not break, because the collision between the egg and the bed sheet lasts over an extended period of time since the bed sheet has some give in it. By increasing the time of the collision, the force of the impact is minimised. Take care to aim at the sheet, because if you miss the sheet, you will definitely break the egg and have to clean up the mess!

Exercise 2.6

A cannon, mass $$\text{500}$$ $$\text{kg}$$, fires a shell, mass $$\text{1}$$ $$\text{kg}$$, horizontally to the right at $$\text{500}$$ $$\text{m·s^{-1}}$$. What is the magnitude and direction of the initial recoil velocity of the canon?

We treat the system as an isolated system and conserve momentum. We choose to the right to be the positive direction. The initial velocity of the system is zero.

Momentum conservation means that $$\vec{p}_{Ti} = \vec{p}_{Tf}$$:

\begin{align*} \vec{p}_{Ti} &= \vec{p}_{Tf} \\ 0 &= m_{cannon}\vec{v}_{cf} + m_{shell}\vec{v}_{sf} \\ 0 &= (500)\vec{v}_{cf} + (1)(+500) \\ -(500)\vec{v}_{cf} &= (1)(+500) \\ \vec{v}_{cf} &= \frac{500}{-500} \\ \vec{v}_{cf} &= -1 \\ \vec{v}_{cf} &= \text{1}\text{ m·s$^{-1}$}~\text{towards the left} \end{align*}

The canon recoils at $$\text{1}$$ $$\text{m·s^{-1}}$$ towards the left.

A trolley of mass $$\text{1}$$ $$\text{kg}$$ is moving with a speed of $$\text{3}$$ $$\text{m·s^{-1}}$$. A block of wood, mass $$\text{0,5}$$ $$\text{kg}$$, is dropped vertically into the trolley. Immediately after the collision, the speed of the trolley and block is $$\text{2}$$ $$\text{m·s^{-1}}$$. By way of calculation, show whether momentum is conserved in the collision.

We calculate the momentum of the system before and after the collision.

Before the collision the velocity of the block is 0. The momentum is:

$$\vec{p}_{i} = m_{1}\vec{v}_1 + m_{2}\vec{v}_2 = 0 + (\text{1})(\text{3}) = \text{3}\text{ kg·m·s^{-1}}$$

After the collision the momentum is:

$$\vec{p}_{f} = (m_1 + m_2)\vec{v} = (\text{1} + \text{0,5})(\text{2}) = \text{3}\text{ kg·m·s^{-1}}$$

Since the momentum before the collision is the same as the momentum after the collision, momentum is conserved.

A $$\text{7 200}$$ $$\text{kg}$$ empty railway truck is stationary. A fertiliser firm loads $$\text{10 800}$$ $$\text{kg}$$ fertiliser into the truck. A second, identical, empty truck is moving at $$\text{10}$$ $$\text{m·s^{-1}}$$ when it collides with the loaded truck.

If the empty truck stops completely immediately after the collision, use a conservation law to calculate the velocity of the loaded truck immediately after the collision.

We will take to the left as positive.

We can use the law of conservation of momentum.

\begin{align*} m_{1}v_{i1} + m_{2}v_{i2} & = m_{1}v_{f1} + m_{2}v_{f2} \\ (\text{18 000})(0) + (\text{7 200})(\text{10}) & = (\text{18 000})v_{f1} + (\text{7 200})(\text{0}) \\ v_{f1} & = \text{4}\text{ m·s$^{-1}$} \text{ to the left} \end{align*}

Calculate the distance that the loaded truck moves after collision, if a constant frictional force of $$\text{24}$$ $$\text{kN}$$ acts on the truck.

From the force we can get the acceleration:

\begin{align*} F & = ma \\ \text{24} \times \text{10}^{\text{3}} & = \text{18 000}a \\ a & = \text{1,33}\text{ m·s$^{-2}$} \end{align*}

Now we can use the equations of motion to find the distance.

\begin{align*} \vec{v}_{f}^2 &= \vec{v}_{i}^2 + 2a\Delta x \\ \text{0} & = \text{4}^{2} + \text{2}(\text{1,33})(\Delta x) \\ \Delta x & = \text{6,02}\text{ m} \text{ to the left} \end{align*}

A child drops a squash ball of mass $$\text{0,05}$$ $$\text{kg}$$. The ball strikes the ground with a velocity of $$\text{4}$$ $$\text{m·s^{-1}}$$ and rebounds with a velocity of $$\text{3}$$ $$\text{m·s^{-1}}$$. Considering only the squash ball, does the law of conservation of momentum apply to this situation? Explain.

The principle of conservation of linear momentum states:

"The total linear momentum of an isolated system is constant. An isolated system has no forces acting on it from the outside."

This means that in an isolated system the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion. Taking downwards as positive.

The momentum before the ball hits the floor is:

\begin{align*} \vec{p}_{down} & = m\vec{v}_i \\ & = (\text{0,05})(+4) \\ & = (\text{0,2}) \\ & = \text{0,2}\text{ kg·m·s$^{-1}$}~\text{downwards} \end{align*}

The momentum after the ball hits the floor is:

\begin{align*} \vec{p}_{up} & = m\vec{v}_f \\ & = (\text{0,05})(-3) \\ & = (\text{0,15}) \\ & = \text{0,15}\text{ kg·m·s$^{-1}$}~\text{upwards} \end{align*}

Since the momentum before the ball hits the floor is not equal to the momentum after the ball hits the floor the law of conservation of momentum does not apply to this situation.

We say that the system is not isolated and that there is a force acting on the ball from outside the system.

A bullet of mass $$\text{50}$$ $$\text{g}$$ travelling horizontally at $$\text{600}$$ $$\text{m·s^{-1}}$$ strikes a stationary wooden block of mass $$\text{2}$$ $$\text{kg}$$ resting on a smooth horizontal surface. The bullet gets stuck in the block.

Name and state the principle which can be applied to find the speed of the block-and-bullet system after the bullet entered the block.

Conservation of momentum

Calculate the speed of the bullet-and-block system immediately after impact.

\begin{align*} m_{1}v_{i1} + m_{2}v_{i2} & = (m_{1} + m_{2})v_{f} \\ (\text{0,05})(600) + (\text{2})(\text{0}) & = (\text{0,05} + \text{2})v_{f} \\ v_{f} & = \text{14,63}\text{ m·s$^{-1}$} \end{align*}

If the time of impact was $$\text{5} \times \text{10}^{-\text{4}}$$ $$\text{s}$$, calculate the force that the bullet exerts on the block during impact.

Calculate the change in momentum of the bullet:

\begin{align*} \Delta \vec{p}_{bullet} &= m(\vec{v}_{f}-\vec{v}_{i}) \\ & = \text{0,05}(\text{14,63} - \text{600}) \\ & = -\text{29,27}\text{ kg·m·s$^{-1}$} \end{align*}

The bullet's change in momentum is equal and opposite to the impulse on the block:

\begin{align*} \text{impulse} & = \vec{F}\Delta t \\ \vec{F} &= \frac{\text{29,27}}{\text{5} \times \text{10}^{-\text{4}}} \\ \vec{F} &= \text{58 540}\text{ N} \end{align*} $$\vec{F} = \text{58 540}\text{ N}$$

Chapter summary (ESCJT)

Presentation: 27JC

• Newton's Second Law: The resultant force acting on a body will cause the body to accelerate in the direction of the resultant force The acceleration of the body is directly proportional to the magnitude of the resultant force and inversely proportional to the mass of the object.

• Newton's Third Law: If body A exerts a force on body B then body B will exert an equal but opposite force on body A.

• Momentum: The momentum of an object is defined as its mass multiplied by its velocity.

• Momentum of a System: The total momentum of a system is the sum of the momenta of each of the objects in the system.

• Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system is constant' or In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)'.

• Elastic collision: both total momentum and total kinetic energy are conserved.

• Inelastic collision: only total momentum is conserved, total kinetic energy is not conserved.

• Impulse:the product of the net force and the time interval for which the force acts.

• Law of Momentum: The applied resultant force acting on an object is equal to the rate of change of the object's momentum and this force is in the direction of the change in momentum.

• Impulse-momentum theorem: the impulse is equal to the change in momentum.

 Physical Quantities Quantity Vector Unit name Unit symbol Mass ($$m$$) - kilogram $$\text{kg}$$ Velocity ($$\vec{v}$$) $$\checkmark$$ metre per second $$\text{m·s^{-1}}$$ Momentum ($$\vec{p}$$) $$\checkmark$$ kilogram metres per second $$\text{kg·m·s^{-1}}$$ Kinetic energy ($$KE$$) - joule $$\text{J}$$ Impulse $$\checkmark$$ newton seconds $$\text{N·s}$$

Table 2.1: Units used in momentum and impulse.

Momentum

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...

1. drop to the ground at zero initial speed.

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

3. be projected upwards at the same initial speed as the first piece.

4. 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?

1. The total linear momentum of the system is not conserved during this collision.

2. The law of conservation of energy does not apply to this system.

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

4. 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?

1. $$\vec{v}$$

2. 2$$\vec{v}$$

3. 3$$\vec{v}$$

4. 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?

1. It will take longer to stop the cricket ball.

2. It will take longer to stop the tennis ball.

3. It will take the same time to stop each of the balls.

4. 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?

1. V

2. 2V

3. 3V

4. 4V

V

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

1. resultant force

2. work

3. power

4. 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?

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

2. $$t$$

3. $$2t$$

4. $$4t$$

$$t$$

[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 ...

1. $$2mv$$

2. $$mv$$

3. $$\frac{1}{2}mv$$

4. $$0mv$$

$$2mv$$

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.

A golf club exerts an average force of $$\text{3}$$ $$\text{kN}$$ on a ball of mass $$\text{0,06}$$ $$\text{kg}$$. If the golf club is in contact with the golf ball for $$\text{5} \times \text{10}^{-\text{4}}$$ $$\text{seconds}$$, calculate

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*}

During a game of hockey, a player strikes a stationary ball of mass $$\text{150}$$ $$\text{g}$$ . The graph below shows how the force of the ball varies with the time.

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{2,7})(\text{2}) + (\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?

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

2. The magnitude of the impulse experienced by the ball is greater than the magnitude of the impulse experienced by the Earth.

3. The speed of the ball before the collision will always be equal to the speed of the ball after the collision.

4. 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