Section 7.4 Elastic And Inelastic Collisions

Section 7.4 Elastic and inelastic collisions

• Conservation of momentum in collisions Energy changes in collisions

• • Explosion and recoil • Apparent non-conservation of momentum

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7.4 Elastic and inelastic collisions (SB p. 166)

Conservation of momentum in collisions Collisions

Elastic collision

rebounds to its original height

Inelastic collision

Law of conservation holds? © Manhattan Press (H.K.) Ltd.

sticks to ground 2

7.4 Elastic and inelastic collisions (SB p. 166)

Elastic collision frictioncompensated runway B

A

plunger

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7.4 Elastic and inelastic collisions (SB p. 166)

Elastic collision Before collision

After collision

• Repeats with 2 and 3 trolleys colliding 1 u stationary trolley A

uA

mA = 2 kg © Manhattan Press (H.K.) Ltd.

mB = 1 kg

mA = 3 kg

mB = 1 kg 4

7.4 Elastic and inelastic collisions (SB p. 167)

Elastic collision Mass / kg

Before collision After collision Initial Total Total Final velocity velocity momentum momentum −1 /ms / m s−1 / kg m s−1 / kg m s−1

mA

mB

uA

uB

mAuA+mBuB

vA

vB

mAvA+mBvB

1

1

0.2

0

0.2

0

0.19

0.19

2

1

0.25

0

0.5

0.085

0.33

0.5

3

1

0.3

0

0.9

0.15

0.44

0.89

Total momentum before collision ≈ Total momentum after collision

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7.4 Elastic and inelastic collisions (SB p. 167)

Data-logging set-up

data-logging interface

Expt 7B Conservation of momentum (data-logging)

motion sensor

cart A

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cart B

motion sensor

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7.4 Elastic and inelastic collisions (SB p. 168)

Data-logging set-up

The v-t graph of A

According to the graphs, is the total momentum conserved?

The v-t graph of B © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 168)

Elastic collision For elastic collision total momentum before collision

total momentum after collision

obeys law of conservation of momentum

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7.4 Elastic and inelastic collisions (SB p. 168)

Inelastic collision

plasticine

trolley A

Expt. 7C Conservation of momentum

trolley B

frictioncompensated runway

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7.4 Elastic and inelastic collisions (SB p. 169)

Inelastic collision Before collision

After collision

• Repeats with 2 trolleys colliding 1 trolley and 1 trolley colliding 2 trolleys

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7.4 Elastic and inelastic collisions (SB p. 169)

Inelastic collision Mass / kg

Before collision After collision Initial Total Total Final velocity velocity momentum momentum -1 /ms -1 -1 /ms / kg m s / kg m s-1

mA

mB

uA

uB

mAuA+mBuB

vA

vB

mAvA+mBvB

1

1

0.23

0

0.23

0.12

0.12

0.24

2

1

0.35

0

0.7

0.23

0.23

0.69

1

2

0.28

0

0.28

0.095

0.095

0.285

Total momentum before collision ≈ Total momentum after collision © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 169)

Inelastic collision For inelastic collision total momentum before collision

total momentum after collision

obeys law of conservation of momentum

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7.4 Elastic and inelastic collisions (SB p. 170)

Example 2: Trolleys A and B are moving towards each other on a smooth horizontal surface as shown in Fig. (a). Take the direction to the right as positive. If trolley A moves at a velocity of 1 m s−1 after the collision (Fig. (b)), find the velocity of trolley B after the collision. Solut ion Fig. (a) Before collision

Fig. (b) After collision © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 170)

Example 2: (Cont) By the law of conservation of momentum, mA uA + mB uB = mA vA + mB vB 2 × 3 + 1 × (−2) = 2 × 1 + 1 × vB ∴ vB = 2 m s−1 The velocity of trolley B after the collision is 2 m s−1 (to the right).

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7.4 Elastic and inelastic collisions (SB p. 170)

Class Practice 2: (a) Refer to Example 2. If trolleys A and B stick together after the collision, find the final velocity of the trolleys. Ans mAuA + mBuB = (mA+ mB) × v wer 2 × 3 + 1 × (−2) = (2 + 1) × v

1 ∴ v = 1.33 m s or 1 3 m s−1 The final velocity of the trolleys is 1.33 m s−1 (to the right). −1

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7.4 Elastic and inelastic collisions (SB p. 171)

Class Practice 2 (Cont): (b)

Ans wer

2 ×3 + 1 × (−2) = 4 kg m s−1 Total momentum before collision = ____________________ Total momentum after collision

2= ×1.33 + 1 × 1.33 = 4 kg m s−1 ____________________

mAvA − mAuA Change in momentum of trolley A = ____________________ −1 2 ×1.33 − 2 × 3 = − 3.33 kg m s ____________________

mBvB − mBuB Change in momentum of trolley B = ____________________ 1×1.33 − 1 × (−2) = 3.33 kg m s−1 ____________________ © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 171)

Energy changes in collisions Collisions

CAL Workshop 2 Elastic and inelastic collisions

elastic Inelastic

Elastic collision

Inelastic collision

• Total momentum is conserved • Will total KE also be conserved?

Go to

Activity 2 © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 172)

Elastic collision On same level, PE = constant Before collision

Mass / kg

mA

mB

2

1

After collision

Before collision Initial Total kinetic velocity energy −1 /J /ms 1 2 1 uA uB mAuA + mBuB2 2 2 0.25

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0

0.0625

After collision Final Total kinetic velocity energy /J / m s− 1 1 2 1 vA vB mAvA + mBvB2 2 2 0.085 0.33

0.0617 18

7.4 Elastic and inelastic collisions (SB p. 173)

Elastic collision For elastic collision Conserved quantities: elastic

• total momentum • total KE

No mechanical energy is lost

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7.4 Elastic and inelastic collisions (SB p. 173)

Inelastic collision Before collision

Mass / kg

mA

mB

1

1

After collision

Before collision After collision Initial Total kinetic Total kinetic common velocity velocity energy energy /J /J / m s− 1 / m s− 1 1 1 2 1 2 uA uB V mAuA + mBuB (mA + mB)v2 2 2 2 0.23

0

0.026 5

0.12

0.014 4

Energy loss = 0.012 1 J © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 174)

Inelastic collision For inelastic collision Conserved quantities: Total momentum only Inelastic

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E K l a t to d e v r e s n o c NOT mechanical energy lost ! 21

7.4 Elastic and inelastic collisions (SB p. 177)

Explosion and recoil Total momentum is conserved in collision: collision

An explosion means the separation of objects which are initially at rest. EXPLOSION Go to

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Is total momentum also Discussion 3 conserved in explosion? 22

7.4 Elastic and inelastic collisions (SB p. 177)

Explosion

Mass / kg

Before explosion Initial Total velocity momentum / m s−1 / kg m s−1

After explosion Final Total velocity momentum / m s−1 / kg m s−1

mA

mB

uA

uB

mAuA + mBuB

vA

vB

mAvA + mBvB

1

1

0

0

0

−0.5

0.5

0

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7.4 Elastic and inelastic collisions (SB p. 178)

Explosion law of conservation of momentum ⇒ valid for explosions

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7.4 Elastic and inelastic collisions (SB p. 178)

Recoil

Total momentum after firing = Total momentum before firing Momentum of the ball + Momentum of the cannon = 0 mv + MV = 0 Go to mv V =− M Discussion 4 © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 179)

Recoil

mv M Mass of cannon (M) >> Mass of ball (m)

For V =

Recoil speed of cannon (V) << Recoil speed of ball (v) Thinking 3 © Manhattan Press (H.K.) Ltd.

V is negative because the cannon moves opposite to ball 26

7.4 Elastic and inelastic collisions (SB p. 179)

Class Practice 3: A bullet of mass 10 g is fired at a speed of 500 m s−1 from a rifle. (a) If the mass of the rifle is 8 kg, find the recoil speed of the rifle.

mv V = ________________________________ M

Ans wer

0.01 × 500 = −0.625 m s−1 = ________________________________ 8 © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 180)

Class Practice 3 (Cont): (b) After improvement, the bullet is fired at a greater speed from the rifle. What would be the change in the total momentum of the system, and the recoil speed of the rifle? Ans wer

Total momentum of the system remains unchanged and is equal to zero. Since the magnitude of the momentum of the bullet increases, the magnitude of the momentum of the rifle also increases. Therefore, the recoil speed of the rifle increases too. © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 180)

Apparent non-conservation of momentum momentum not conserved?

NO

Momentum of the boy + Momentum of Earth = constant © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 180)

Class Practice 4: When a bullet is fired towards a wall and embedded in it, its velocity becomes zero. The momentum of the bullet becomes zero too. Does it mean that the law of conservation of momentum is invalid here? Explain briefly. Ans wer When the bullet is embedded into the wall, its momentum is decreased. On the other hand, the momentum of the earth is increased due to the force acting on the wall by the bullet. Since the movement of the earth is so small, it is unnoticeable. As a result, the total momentum of the bullet and the earth remains unchanged. © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 181)

Useful Website

Elastic and Inelastic Collision (http://www.walter-fendt.de/ph14e/collision .htm) Shocked Air Track (http://www.edp.ust.hk/physics/explore /dswmedia/airtrack.htm) Automotive Coalition for Traffic Safety. Inc. (http://www.actsinc.org) © Manhattan Press (H.K.) Ltd.

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7.4 Elastic and inelastic collisions (SB p. 180)

Mind Map

how to measure motion?

7.1 Momentum

7.2 Change in momentum and net force

no external force is acted on the system

a net force acts on an object for a period of time

Momentum

• resulted in a change in momentum • the object receives an impulse • according to Newton’s 2nd law, impulsive force = rate of change in momentum 7.4 Elastic and inelastic collisions

p = mv

7.3 Law of conservation of momentum in collision, momentum is conserved momentum is conserved in all collisions when KE is conserved

elastic collision

when KE is not inelastic collision conserved © Manhattan Press (H.K.) Ltd.

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The The End End

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7.4 Elastic and inelastic collisions (SB p. 171)

Activity 2: Energy change in collisions elastic ball

ping-pong ball plasticine

1. Drop an elastic ball at a certain height above the ground. Describe the motion of the ball. State the energy change during the process. Ans The ball rebounds almost to the original height. During the wer downward motion, the PE of the ball changes to KE. During the upward motion, the KE changes to PE. The energy loss is negligible.

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7.4 Elastic and inelastic collisions (SB p. 171)

Activity 2 (Cont): Energy change in collisions

2. Drop a ping-pong ball at the same height. Describe the motion of the ball. State the Ans energy change during the process.

The ping-pong ball rebounds to a smaller height. The energy wer changes are similar to those in step 1. But there is a considerable amount of energy loss as heat and sound during the collision.

3.Drop a piece of plasticine at the same height. Describe its motion. State the energy change Ans during the process. The plasticine sticks to the ground. All the potential energy of the plasticine changes to the internal energy, heat and sound. © Manhattan Press (H.K.) Ltd.

wer

Return to

Text 35

7.4 Elastic and inelastic collisions (SB p. 177)

Discussion 3:

An explosion means the separation of objects which are initially at rest. For example, the cork ejects from the bottle during an explosion. Does the law of conservation of momentum still hold in Ans an explosion? werejects Yes, it holds. When the cork from the bottle, the bottle moves slightly backwards at the same time. The total momentum of them is zero before and after the explosion. Since we are holding the bottle, the mass of the bottle becomes very large, the movement of the bottle is very small. © Manhattan Press (H.K.) Ltd.

Return to

Text

36

7.4 Elastic and inelastic collisions (SB p. 178)

Discussion 4

Have you ever seen cannons or rifles move backwards when they are fired? Do you know why?

A pistol Return to

A cannon A rifle © Manhattan Press (H.K.) Ltd.

Text 37

7.4 Elastic and inelastic collisions (SB p. 179)

Thinking 3:

1. Give a pair of roller-skates to a student to wear. Ask the student to hold a basketball and then throw it forwards. Observe what happens to him and explain briefly. Ans momentum,wer the

According to the conservation of student moves backwards when he throws the ball forwards.

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7.4 Elastic and inelastic collisions (SB p. 179)

Thinking 3 (Cont):

2. When water is ejected from the water sprinkler, the nozzle rotates. Explain briefly. Ans wer

Return to

Text When the water is ejected, the reaction makes the nozzle rotate. © Manhattan Press (H.K.) Ltd.

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