Physics P Worksheet 9.2 Conservation of Momentum

Worksheet 9.2 Conservation of Momentum

1. Mya and Kengo are both at rest and facing each other on roller skates. Mya has a mass of 65 kg and Kengo

has a mass of 40 kg. When they push against each other Mya moves at a speed of 6 m/s. a) How fast does Kengo move? b) How much work did they do pushing against each other?

2. A loaded freight car whose mass is 16 000 kg rolls along a level track at a speed of 1.5 m/s toward an empty freight car. When the two cars collide, they couple and proceed at 1 m/s. a) Determine the mass of the empty freight car. b) What are the initial and final kinetic energies of the system and what percent of the energy was lost

(energy lost / initial energy). 3. A 6–kg mass moving at 10 m/s has a head-on collision with a 4–kg mass that is initially at rest. Following the

collision, the 6 kg mass moves at a speed of 2 m/s. a) Find the final speed of the 4–kg mass. b) Is this collision elastic, inelastic, or something in between? Justify your answer.

4. A baseball (m = 0.149 kg) is pitched at a speed of 36 m/s. The batter swings a bat (m = 0.85 kg) at a speed of 15 m/s. After the collision, the baseball is moving in straight toward the pitcher at a speed of 42 m/s. a) What is the speed of the bat after the collision? b) How much heat was generated in this collision?

5. A 3.2–kg handgun fires a 35–g bullet. The bullet leaves the handgun with a velocity of 280 m/s. What is the velocity of recoil of the handgun?

6. A 227Th nucleus (m = 227 u, don’t worry about the units, they will cancel out) decays into a 223Ra nucleus by α emission (an α particle is the same as a 4He nucleus). When a 227Th nucleus at rest decays, the a particle has a final speed of 1.4 × 107 m/s. What is the final speed of the recoiling 223Ra nucleus?

7. An open railroad with a mass of 12 000 kg roll at a speed of 3 m/s when it starts raining. What is the velocity of the car when 6 000 kg of water has accumulated in it?

8. A football running back with a mass of 110 kg runs at 4 m/s. A stationary free safety with a mass of 85 kg grabs him and the two continue to move together. What is their new combined speed?

9. A VW Beetle with a mass of 850 kg travels at a speed of 25 m/s when it has a head–on collision with a pick–up with a mass of 2080 kg. Both vehicles come to a rest as a result of the collision. a) What was the speed of the pick–up? b) How much kinetic energy was lost in the collision?

10. An astronaut with a mass of 140 kg working in space accidently pushes off from his spacecraft and begins drifting away at a speed of 0.5 m/s. To get back he throws his 5–kg tool bag at a speed of 16 m/s. At what speed will he be moving back toward the spacecraft?

Physics P Worksheet 9.2 Conservation of Momentum 1a. Before After

p =mv pbefore = 0 pafter = 0

pM +pK = 0mMvM +mKvK = 0

vK = −mMvM

mK

vK = −65$kg( ) 6$m/s( )

40$kgvK = −9.8$m/s

Kengo moves backward at 9.8 m/s.

1b. W = ΔKEM +ΔKEK

W = 12mMvM

2 + 12mKvK

2

W = 12 65%kg( ) 6%m/s( )2 + 1

2 40%kg( ) 9.8%m/s( )2W =1170%J%+%1900%JW = 3100%J

They do 3100 J of work.

2a. Before After

p =mvp = 16#000#kg( ) 1.5#m/s( )p = 24#000#kg ⋅m/s

p =mvp = 0

pbefore = 24#000#kg ⋅m/s pafter = 24#000#kg ⋅m/s

p =mv

m = pv

m = 24#000#kg ⋅m/s1#m/s

m = 24#000#kg

The mass of both cars is 24 000 kg. The second car must have a mass of 8 000 kg.

Physics P Worksheet 9.2 Conservation of Momentum 2b. Before After

KEbefore =12mvbefore

2

KEbefore =12 16$000$kg( ) 1.5$m/s( )2

KEbefore =18$000$J

KEafter =12mvafter

2

KEafter =12 24$000$kg( ) 1$m/s( )2

KEafter =12$000$J

energy&lostinitial&energy

= 18&000&J&1&12&000&J18&000&J

= 0.33=&33%

KEbefore = 18 000 J, KEafter = 12 000 J, resulting in a 33% loss.

3a. Before After

p1 =m1v1

p1 = 6#kg( ) 10#m/s( )p1 = 60#kg ⋅m/s

p2 =m2v2

p2 = 4#kg( ) 0#m/s( )p2 = 0

pbefore = 60#kg ⋅m/s pafter = 60#kg ⋅m/s

p1 =m1v1

p1 = 6#kg( ) 2#m/s( )p1 =12#kg ⋅m/s

p2 = 48#kg ⋅m/s

v2 =p2

m2

v2 =48#kg ⋅m/s

4#kgv2 =12#m/s

After the collision the 4 kg mass is moving at 12 m/s.

3b. Before After

KE1 =12mv1

2

KE1 =12 6$kg( ) 10$m/s( )2

KE1 = 300$J

KE2 =12mv2

2

KE2 =12 4$kg( ) 0$m/s( )2

KE2 = 0$J

KE1 =12mv1

2

KE1 =12 6$kg( ) 2$m/s( )2

KE1 =12$J

KE2 =12mv2

2

KE2 =12 4$kg( ) 12$m/s( )2

KE2 = 288$J

KEbefore = 300#J KEafter =12#J#+#288#J

KEafter = 300#J

Since the total kinetic energy before the collision is equal to the total kinetic energy after the collision, the collision is elastic.

Physics P Worksheet 9.2 Conservation of Momentum 4a. Before After

pball =mballvball

pball = 0.149&kg( ) )36&m/s( )pball = )5.364&kg ⋅m/s

pbat =mbatvbat

pbat = 0.85%kg( ) 15%m/s( )pbat =12.75%kg ⋅m/s

pbefore = 7.386&kg ⋅m/s pafter = 7.386&kg ⋅m/s

pball =mballvball

pball = 0.149&kg( ) 42&m/s( )pball = 6.258&kg ⋅m/s

pbat =1.128%kg ⋅m/s

vbat =pbat

mbat

vbat =1.128%kg ⋅m/s

0.85%kgvbat =1.3%m/s

The final speed of the bat is 1.3 m/s.

4b. Before After

KEball =

12mvball

2

KEball =12 0.149'kg( ) 36'm/s( )2

KEball = 96.6'J

KEbat =12mvbat

2

KEbat =12 0.85'kg( ) 15'm/s( )2

KEbat = 95.6'J

KEball =

12mvball

2

KEball =12 0.149'kg( ) 42'm/s( )2

KEball =131.4'J

KEbat =12mvbat

2

KEbat =12 0.85'kg( ) 1.3'm/s( )2

KEbat = 0.72'J

KEbefore = 96.6$J$+$95.6$J

KEbefore =192.2$J

KEafter =131.4%J%+%0.72%J

KEafter =132.1%J

Q = ΔKEQ =192.2%J%'%131.1%JQ = 61.1%J

There was 61.1 J of heat generated in the collision.

5. Before After p =mv pbefore = 0 pafter = 0

pb +pg = 0

mbvb +mgvg = 0

vg = −mbvb

mg

vg = −0.035%kg( ) 280%m/s( )

3.2%kgvg = −3.1%m/s

The gun moves backward at 3.1 m/s.

Physics P Worksheet 9.2 Conservation of Momentum 6. Before After

p =mv pbefore = 0 pafter = 0

pRa +pHe = 0mRavRa +mHevHe = 0

vRa = −mHevHe

mTh

vRa = −4#u( ) 1.4# × #107 #m/s( )

223#uvRa = −2.5# × #105 #m/s

The radium nucleus recoils backward at 2.5 × 105 m/s.

7. Before After

pcar =mcarvcar

pcar = 12#000#kg( ) 3#m/s( )pcar = 36#000#kg ⋅m/s

prain =mrainvrain

prain = 0

pbefore = 36#000#kg ⋅m/s pafter = 36#000#kg ⋅m/s

p =mv

v = pm

v = 24#000#kg ⋅m/s12#000#kg#+#6#000#kg

v = 2#m/s

The velocity after the car is full of rain is 2 m/s.

8. Before After

prb =mrbvrb

prb = 110#kg( ) 4#m/s( )prb = 440#kg ⋅m/s

pfs =mfsvfs

pfs = 0

pbefore = 440#kg ⋅m/s pafter = 440#kg ⋅m/s

p =mv

v = pm

v = 440#kg ⋅m/s110#kg#+#85#kg

v = 2.3#m/s

The velocity after the collision is 2.3 m/s.

Physics P Worksheet 9.2 Conservation of Momentum 9a. Before After

p =mv pbefore = 0 pafter = 0

pVW +ptruck = 0mVWvVW +mtruckvtruck = 0

vtruck = −mVWvVW

mtruck

vtruck = −850$kg( ) 25$m/s( )

2080$kgvtruck = −10.2$m/s

The truck was moving in the opposite direction at 10.2 m/s.

9b. Before After KEVW = 1

2mVWvVW2

KEVW = 12 800%kg( ) 25%m/s( )2

KEVW = 266%000%J

KEtruck =12mtruckvtruck

2

KEtruck =12 2080%kg( ) 10.2%m/s( )2

KEtruck =109%000%J

KEafter = 0

KEbefore = 266#000#J#+#109#000#J

KEbefore = 374#000#J

ΔKE = KEafter !+!KEbefore

ΔKE = 0!J−374!000!JΔKE = −374!000!J

There were 374 000 J of kinetic energy lost.

10. Before After

p =mvpbefore = 140$kg$+$5$kg( ) 0.5$m/s( )pbefore = 72.5$kg ⋅m/s

pafter = 7.386&kg ⋅m/s

ptool =mtoolvtool

ptool = 5"kg( ) 16"m/s( )ptool = 80"kg ⋅m/s

pastronaut = 72.5%kg ⋅m/s%+%80%kg ⋅m/s

pastronaut = +7.5%kg ⋅m/s

vastronaut =pastronaut

mastronaut

vastronaut =+7.5%kg ⋅m/s140%kg

vastronaut = +0.05%m/s

The astronaut moves backwards at a speed of 0.05 m/s.

Physics P Worksheet 9.2 Conservation of Momentum