momentumbegining... · large that we may be able to ignore the much smaller external ... apply the...

189

Momentum

CHAPTER 7

LEARNING OBJECTIVES

The learning objectives for this chapter are listed below. By ticking each objective when you have met it, you can track your progress over the course.

Identify concepts from the list below (‘Achievement’ criterion one):

momentum

the conservation of momentum in isolated systems

the change in momentum of an object in one dimension

impulse and force.

Describe situations involving concepts from the list below (‘Achievement’/‘Merit’ criterion one):

momentum



impulse and force.

Explain situations involving concepts from the list below (‘Merit’/‘Excellence’ criterion one):

momentum



impulse and force.

Solve problems involving concepts from the list below (‘Achievement’/‘Merit’/‘Excellence’ criterion two):

momentum



impulse and force.

190 Beginning Physics Workbook

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K MOMENTUMIf a mass m moves with velocity v, then its momentum p is given by:

p = mv

Momentum has the unit kg m s–1, or kilogram metre per second. Momentum is a vector and its direction is the direction of the velocity.

m

p = mv

v

Example

2 kg1 m s–1

has the same momentum as 1 kg2 m s–1

but a different momentum from 2 kg1 m s–1

.

Momentum may be thought of as quantity of motion. The more momentum something has, the harder it is to stop.

Example

If two objects have the same mass but the second is moving twice as fast then it will be twice as hard to stop. This is because it has twice the ‘quantity of motion’, twice the momentum.

Similarly, if two objects move with the same speed but the second has three times the mass then it will be three times as hard to stop. This is because it has three times the ‘quantity of motion’, three times the momentum.

Conservation of momentumWhen two bodies A and B collide, each exerts a force on the other. From Newton’s third law, we know that these two forces are equal in size and opposite in direction. Usually the collision will last only a short time and, for this short time, the two forces will be large.

For two colliding bodies A and B, if the collision forces are the only forces that need to be considered, then we fi nd that total momentum is conserved.

This means that if we add up the momentum of A and B before the collision, and then we add up the momentum of A and B after the collision, we would fi nd the two answers are the same.

If, in the collision, A loses 5 kg m s–1 of momentum then we would fi nd that B gains 5 kg m s–1 of momentum.

http://faraday.physics.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/AirTrack/AirTrack.html

http://www.walter-fendt.de/ph14e/ncradle.htm

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Chapter 7 Momentum 191

Kinetic energy need not be conserved in such collisions but momentum always is. When kinetic energy is conserved in a collision, we say that the collision is elastic.

A pair or a collection of masses (i.e. a system) on which no external forces act is called an isolated system. Most of the situations we will deal with in NCEA Level 2 Physics will be collisions or explosions within an isolated system.

Example

A 3 kg body moves to the right with speed 6 m s–1 while a 5 kg body moves to the left with speed 2 m s–1. They collide and stick together.a What is their speed after the collision?

After the collision there is effectively just one body of mass 8 kg. We will call its speed v:

Before:

After:

Sign convention:

Since momentum is conserved:momentum before = momentum after 3 × 6 + 5 × –2 = 8 × v 18 – 10 = 8v 8 = 8v v = 1 m s–1

The answer is positive so the direction is to the right.

b Is the collision elastic? Before the collision:

kinetic energy, = =kE mv12

12

3 62 2× × +

64 J

12

5 22× ×

=

After the collision:

Ek = = J12

8 1 42× ×

We see that kinetic energy is not conserved and the collision is therefore not elastic. In fact, collisions in which objects stick together are never elastic. The kinetic energy that has been lost will have turned into other forms of energy such as heat and sound.

5 kg2 m s–1

8 kgv m s–1

3 kg6 m s–1

+−

http://www.walter-fendt.de/ph14e/collision.htm


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K Example

On icy winter roads, a 1500 kg car travelling at 21 m s–1 collides with an 1800 kg van travelling at 15 m s–1 in the opposite direction. The two vehicles lock together as a crumpled wreck and move off with speed v. Calculate v.

The total initial momentum = piA + piB

= 4500 kg m s–1

The fi nal momentum = 4500 kg m s–1

pf = mf vf

4500 = (1500 + 1800) × vf

And since this fi nal velocity value is positive, we can see that the wreck will move off in the original direction that the 1500 kg car was travelling in.

1500 kg 21 m s–1 1800 kg15 m s–1

A B

The initial momenta:(taking the positive direction as → )

piA = 1500 × 21= 31 500 kg m s–1

piB = 1800 × –15= –27 000 kg m s–1

vf = = m s 2 sig figs45003300

1 4 1. ( )−

Example

An ice-hockey puck of mass 0.80 kg moving at 3.5 m s–1 hits the side of a second puck that was initially at rest. The mass of the second puck is 0.70 kg.

After the collision, the 0.80 kg puck moves off at 2.8 m s–1 at right-angles to its original direction.

Find the velocity of the 0.70 kg puck immediately after the collision.To tackle this problem, we need to use conservation of momentum. This is an

isolated system. The total change in momentum of the system is zero. This means that the change in momentum of the 0.80 kg puck is equal and opposite to the change in momentum of the 0.70 kg puck.

The change in momentum of the 0.80 kg puck is:

∆ p = pf – pi

0.8 kg3.5 m s–1

0.7 kgBefore the collision:

0.8 kg

2.8 m s–1

0.7 kg

After the collision:

v = ?

∆ means ‘changes in’, i.e. ‘fi nal – initial’.

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Momentum is a vector, so:

∆p = 0.8 × 2.8 kg m s–1 – 0.8 × 3.5 kg m s–1

∆p = 2.24 kg m s–1 + 2.8 kg m s–1⇔

To subtract vectors, we ‘add the opposite’.

∆ +p = ( . ) ( . )2 24 2 82 2

= 3.6 kg ms–1 (2 sig fi gs)

The direction is such that θ = tan2.242.8

= 391−

°

∆p

(2.8)

(2.24)

This means that the change in momentum of the 0.70 kg puck is:

39°

3.6 kg m s–1

and ∆ p = pf – pi, but pi = 0 since this puck was initially at rest, so:

∆ p = pf = 3.6 kg m s–1

vpm

= = = m s3 6

0 705 1 1.

.. −

Example

A 4.9 kg block of wood is suspended by a light long string. A bullet of mass 0.10 kg is fi red horizontally into it with a speed of 100 m s–1. How high will the block swing? Take g = 10 m s–2.

4.9 kg5.0 kg

100 m s–1

0.10 kgh

Note that the bullet’s kinetic energy doesn’t all get converted into the potential energy that the block has at the top of its swing. This is because there are losses to friction as the bullet slows down in the wood. Most of the bullet’s kinetic energy actually turns into heat. The collision between the bullet and the block is inelastic.

http://www.youtube.com/watch?v=qHbcsLp8dDI


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IMPULSEThe impulse of a force is defi ned by:

impulse = F × t (the force multiplied by the time for which the force acts)

Since =

f i

f i

F ma

Fm v v

tFt mv mv

⇒ =−

⇒ = −

( )

In words:

impulse = change in momentum The units of impulse are N s or kg m s–1.

If two objects have the same mass but the second is moving twice as fast as the fi rst, then it will be twice as hard to stop. This means that the impulse required would be twice as large for the second object. We would need either:• a force twice as big acting for the same length of time, or• the same size force acting for twice as long.

We need to fi nd the speed of the wood after the bullet is lodged in it. For this, we use conservation of momentum.

Before: After:

100 m s–1 0 m s–1 v m s–1

0.10 4.9 5.0

p p

vi f=

× + × = ×0 10 100 4 9 0 5 0. . . (recalll p mv=⇒ = −

)

.v m s2 0 1

For the maximum height the block reaches, we use the fact that the kinetic energy the block + bullet have after the collision turns into potential energy as the block swings up:

E E

mv mgh

v gh

k ploss gain=

=

=

×

12

12

0 5 2

2

2

. ...

.

0 102 0 10

0 20

2

m

===

h

h

h

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A PROOF THAT MOMENTUM IS CONSERVEDSuppose two bodies A and B collide. Then by Newton’s third law, the force on body A (FA) will be equal and opposite to the force on body B (FB) provided no external unbalanced forces act. The two forces mentioned must also act for the same time (t).Thus we have:

FA = –FB

⇒ FA × t = –FB × t⇒ impulse A = –impulse B⇒ change in momentum of A = –change in momentum of B

Therefore, if A loses a certain amount of momentum then B will gain that same amount of momentum. The total momentum will therefore stay the same. It is conserved.

If no external unbalanced forces act in a collision, then momentum is conserved.

If there are external forces acting, i.e. if the system is not isolated, then momentum need not be conserved. This is because with external forces, the force on A need not be equal and opposite to the force on B.

In a collision or an explosion, the force each mass exerts on the other is often very large, so large that we may be able to ignore the much smaller external forces. In this case we can still apply the conservation of momentum principle.

Practical work

A) COLLISIONSTie two 0.5 kg trolleys together with a long piece of elastic.

Hold the trolleys at least 2 m apart, marking their starting positions.

Release the trolleys at the same time.Make two measurements:

• the position where the trolleys collide• the time taken from release to collision.

Record your results using the table below and do the calculations. Add masses to the trolleys to vary the ratio of m1 : m2.

m2m1

m1 m2

x1 x2Release

Trolleys collide

http://www.ngsir.netfirms.com/englishhtm/DropABrick.htm


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It is worth doing three to four collisions for each set-up, and recording the average x1, x2 and t values.

m1

(kg)m 2

(kg)x1

(m)x2

(m)t

(s)v1

(m s–1)v2

(m s–1)p1

(kg m s–1)p2

(kg m s–1)

0.5 0.5

0.5 1.0

0.5 1.5

1.0 0.5

1.0 1.0

1.0 1.5

Note: you may have different values for m1 and m2 than those shown in the table, depending on the equipment you use.

1 Comment on the relationship between p1 and p2.

2 Explain how the elastic provides the same force on each of the trolleys at release.

3 How does this experiment demonstrate Newton’s third law and the conservation of momentum?

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B) DROPPING A MASS ONTO A TROLLEY

Use a motion sensor (or ticker-tape, or a stopwatch and tape measure) to measure the speed of a trolley just before and just after a mass is dropped onto it. Repeat for a few different values of m1 and m2.

Drop gently from asclose as possible

m1v1

Initial set-up:

Final set-up:

m1v2

m2

m2

Record your data in the table below and do the calculations.

Trial m1 (kg)

m2 (kg)

v1 (kg)

v2

(m s–1)m1v1 = pi (kg m s–1)

(m1 + m2) v2 = pf

(kg m s–1)

a 0.5 0.5

b 1.0 0.5

c 1.5 0.5

Note: you may have different values for m1 and m2 than those shown in the table, depending on the equipment you use.

1 Comment on the relationship between pi and pf.


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K 2 Use the motion sensor data to graph velocity vs time for each of the trials a, b and c. On each of your graphs, show when the mass was dropped onto the trolley and comment on the shape of the graph.

Comments:

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MQuestions for discussion

1 Explain the effect on the momentum of the planet Earth if everyone in the Southern Hemisphere jumped up in the air at exactly the same time.

2 If you were standing in the middle of a completely frictionless ice rink with smooth, slippery shoes on – how could you get yourself to the other side?

3 What is the best way for an out-of-control ice-skater to stop?a falling overb crashing into someone travelling the same wayc crashing into someone travelling towards them (‘head on’ collision) d crashing into the wall.Explain why.

4 In a dispute between two astronauts in outer space, angry astronaut A pulls a gun out and shoots startled astronaut B. Can the bullet kill B?


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Problems and exercises

1 A 65 kg skater travelling at 11 m s–1 crashes into a brick wall. The skater is stopped by the wall in 0.67 s.a Calculate the deceleration of the skater.

b Find the force on the skater due to the wall.

c What is the force on the wall due to the skater?

2 A 0.500 kg lab trolley (A) collides with a stationary 0.500 kg trolley (B).

BA

The force of impact on A due to B was 20 N. What was the force on B due to A?

3 A 100 g ball bounces off a wall with no change in speed, as shown below.

vi = 12 m s–1

vf = 12 m s–1

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Find the change in momentum of the ball.

4 Find the momentum of:a a 1.6 kg brick falling, at the instant it is travelling at 20 m s–1

b a radioactive particle of mass 9.11 × 10–31 kg moving at 75% of c (the speed of light) – ignoring relativistic effects.

c a 900 kg car travelling at 28 m s–1.

Problems and exercises

1 A 65 kg skater travelling at 11 m s–1 crashes into a brick wall. The skater is stopped by the wall in 0.67 s.a Calculate the deceleration of the skater.

b Find the force on the skater due to the wall.

c What is the force on the wall due to the skater?

2 A 0.500 kg lab trolley (A) collides with a stationary 0.500 kg trolley (B).

BA

The force of impact on A due to B was 20 N. What was the force on B due to A?

3 A 100 g ball bounces off a wall with no change in speed, as shown below.

vi = 12 m s–1

vf = 12 m s–1


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K 5 An air-track glider of mass 0.25 kg moving at 2.5 m s–1 collides with a stationary glider. Immediately after the collision, the gliders stick together moving at 1.0 m s–1. Find the mass of the second glider.

0.25 kg

6 Two gliders of equal mass on an air track travel towards each other with speeds of 7.5 m s–1 and 2.5 m s–1. When they collide, they stick together. At what speed do they travel after the collision?

7 A 1400 kg car travels East at a speed of 20 m s–1 and collides head-on with a 700 kg car which travels West at a speed of 15 m s–1. After the collision, the 1400 kg car is moving East with a speed of 5 m s–1. Use the sign convention that East is positive.a Find the speed of the 700 kg car after the collision.

b The change in velocity of the 1400 kg car is –10 m s–1. What is the change in velocity of the 700 kg car?

c Use conservation of momentum to explain why the change in velocity is bigger for the less massive car.

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d Use what you have found out above to explain why the occupants of the more massive car may suffer less injury than the occupants of the less massive car in this collision.

8 A 0.02 kg bullet travelling at 200 m s–1 hits a 12 kg hanging sandbag. Calculate the initial speed of the sandbag with the embedded bullet.

9 An ice-hockey player of mass 95 kg travelling at 8.4 m s–1 collides head-on with another player (but of mass 80 kg). The collision stops both players instantly. At what speed was the 80 kg player travelling before the collision?

10 A bullet of mass 30 g is fi red from a 4.6 kg gun at a speed of 450 m s–1. Find the recoil speed of the gun.


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K 11 A 10.0 kg artillery shell moving at 90 m s–1 breaks into two parts with equal mass and equal speeds.Find the speed of the parts after the explosion.

35 °

35 °

vf

vf

After explosion:

10.0 kg

90 m s–1

Before explosion:

12 A 10 kg bomb travelling at 65 m s–1 breaks apart with no overall loss of mass as shown below:

Explosion40°

120 m s–1

50°

4 kg

10 kg A

B

a Find the momentum of piece B.

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b What is the mass of piece B?

c Calculate the speed of piece B.

13 A 1.2 kg object moving at 4.0 m s–1 due East collides with a 3.0 kg object moving at 2.0 m s–1 due North. Following the collision, the two objects stick together.a Sketch a diagram of the situation before the collision.

b Assuming this is an isolated system, fi nd the momentum of the two objects after the collision.

c Find the velocity of the two objects after the collision.

14 A 1.5 × 103 kg car (including all contents), travelling at 25 m s–1, collides with a 2600 kg mini-bus in a head-on smash. To the great surprise of everyone watching, both vehicles are stopped by the impact (they lock together with a fi nal velocity of zero!).a Sketch a diagram of the situation before and after the smash.


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K b What is the total momentum of the system after the smash?

c What is the total momentum of the system before the smash? Explain why.

d Calculate the initial speed of the mini-bus (prior to the smash). Show all your working.

15 A 1.5 kg bomb travelling at 200 m s–1 due East explodes into two equal parts as shown below:

25°200 m s–1

25°

0.75 kg

1.5 kg

0.75 kg

v

v

AfterBefore

Each of the exploded parts moves off with the same speed but in different directions.a What is the total fi nal momentum of the system?

b Sketch a vector diagram (to scale) to fi nd the fi nal momentum of each of the two parts.

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c Using your momentum vector diagram above, fi nd the value of the fi nal speed v of each of the exploded parts.

16 A 50 g golf ball is driven with a speed of 70 m s–1. The impact with the golf club lasts l.2 × l0–3 s. Find:a the change in momentum of the golf ball

b the impulse on the golf ball

c the average force acting on the golf ball during the impact

d the average force acting on the golf club during the impact.

17 Trolley A collides with trolley B. The force on B due to A is 2.50 N.

BA1.0 m s–1

1.0 kg 2.0 kg1.0 m s–1

a What is the force on A due to B?

b State the principle that you used to answer a.

c Calculate the deceleration of trolley B due to this collision. (Assume no friction.) Give your answer in a sentence with correct SI units and sig fi gs.


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K 18 A train collides with an empty cattle truck and keeps moving along the rail tracks with the truck attached. A velocity–time graph shows the momentum of the train (and the truck).

a The impact occurred between t = 6 s and t = 7 s. Calculate the deceleration of the train during this second.

b The total time elapsed from when the train driver fi rst saw the truck until the train fi nally slid to a halt was 16 s. Calculate the distance covered by the train over this time.

c Write a description of a possible journey the train could make to give this graph.

19 A 70 kg ice-hockey player collides with a 50 kg player as shown.a Find the total momentum of the two players

before the collision.

b During the collision, all of the momentum is transferred to the 50 kg player. Find the change in momentum of the 50 kg player.

10

20

2 4 6 8 10 12 14 16t (s)

The crashv

(m s–1)

70 kg

6 m s–1

50 kg

4 m s–1

momentumbegining... · large that we may be able to ignore the much smaller external ... apply the...

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