worksheet 9.2 conservation of momentum - trunnell's...
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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