1 1 2 2 1 1 2 2 1 1 1 1 2 2 2 2 1 1 2 2 2 1 1 1 ( ) (0.18 kg)( 2.5 m/s) (0.25 kg)(1.7 m/s ( 0.10 m/s)) 0.18 kg 0 m/s m v m v m v m v m v m v m v m v m v m v v v m v ′ ′ + = + ′= + − ′ ′ + − ′ = − + − − = ′ =
The velocity of the 0.18-kg ball after the collision is 0 m/s.
Applying Inquiry Skills
11. Set up the carts as shown in the text on a track. Attach a spark timing recorder to each cart to record location for each time interval. Turn on the timers, and release the carts. Analyze each tape close to the collision to determine the speed of each cart. Mass each cart and calculate the momentum of each cart after the collision. Check to see if each one adds up to zero (as it should).
12. (a) Since the car and SUV came to an immediate halt at the location of the crash, the total momentum of the system was zero, before and after the collision. That would mean that the total momentum before and after would have the same magnitude. C C S S S C C C C S C C C S C S (but 2 ) (2 ) 2 2 m v m v m m m v m v m v m v v v = = = = =
This data indicates the car was travelling at twice the speed of the SUV.
(b) If both drivers had the same speed, there would have been momentum after the collision in the direction of the original motion of the SUV.
5.3 ELASTIC AND INELASTIC COLLISIONS
Try This Activity: Newton’s Cradle
(Page 248)
(a) Each collision is successive. Assuming an elastic collision,
1 1 2 2 1 1 2 2 2
1 1 2
1 1 2
2 1 1
(but 0 and all masses are equal) 0 (Equation 1) m v m v m v m v v mv mv mv v v v v v v ′ ′ + = + = ′ ′ + = + ′ ′ = + ′ = − ′ and 2 2 2 2 1 1 1 1 1 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 1 1
(but 0 and all masses are equal) 0 (Equation 2) m v m v m v m v v mv mv mv v v v v v v ′ ′ + = + = ′ ′ + = + ′ ′ = + ′ = − ′ Substitute Equation 1 into Equation 2:
2 2 2 1 1 1 1 2 2 2 2 1 1 1 1 1 1 2 1 1 1 ( ) 2 2 2 0 ( ) 0 v v v v v v v v v v v v v v v v ′ ′ − = − ′ ′ ′ − + = − ′ ′ / −/ = ′ ′ − =
Substituting back into Equation 1 gives: 2 1 1 1 2 1 0 v v v v v v ′= − ′ = − ′ =
Each sphere collision will leave the previous one stationary as the next one moves on. Only one sphere can move out from the end.
(b) Similarly, only one sphere can end up with all of the kinetic energy.
(c) According to the above calculations, each sphere will “pass” its momentum onto the next. When the last sphere receives the momentum, and therefore, speed, it will rise to the same vertical height as the original sphere was dropped.
(d) If two spheres are used, then two spheres will move out from the other side. If three spheres are used, then three spheres will move out from the other side.
PRACTICE
(Page 248)
Understanding Concepts
1. All of the original kinetic energy is transformed to other forms if both objects come to rest after the collision.
2. Yes, we can conclude the collision is completely inelastic. After a “hit-and-stick” collision, no energy is stored as elastic potential to be returned to either object at the end of the collision.
3. A head on collision is very dangerous because of the high relative velocity between the vehicles and the large (and rapid) change in speed for each one. These large accelerations produce large forces that are capable of inflicting serious damage on a human body.
Applying Inquiry Skills
4. If the ball tends to return to shape quickly when squeezed, it will have a more elastic collision than one that returns to its original shape more slowly.
5. (a)
(b)
Making Connections
6. (a) The impact force is reduced by the soft interior because it takes the same impulse, and causes the interaction to last longer. The larger time interval of collision means that a smaller force is applied. A hard interior would have a short duration of collision, and a higher force.
(b) If a helmet does not fit properly, the force applied to the head is not evenly distributed to the whole head. The smaller distribution area means the force would be applied to a smaller area, increasing the pressure to that portion of the head. (c) Once a helmet is involved in a collision, it should be replaced. One way a helmet is designed to reduce injury is to
absorb some of the impact by breaking. This is only able to help the wearer one time. In future collisions, the safety of the wearer could be compromised.
7. Some possible answers: Flexible
•
absorb energy•
reduce deceleration of train (and passengers) Rigid•
prevent car from collapsing and injuring passengers located at either end•
less likely to have debris flying about the interior (due to crumpling car)PRACTICE
(Pages 251–252)
Understanding Concepts
8. The larger truck would have the larger momentum.
2 K 2 2 2 2 K K 2 2 ( ) 2 2 2 mv E m v m mv m p E m p mE = = = = =
psmall = 2msmall K smallE and plarge= 2mlargeEK large Since ,EK small=EK large
psmall = 2msmall KE and plarge= 2mlarge KE The mass of the larger vehicle is larger, and the momentum will be too.
9. (a) All objects have mass. If it also has a velocity, then it will have both momentum and kinetic energy.
(b) For an isolated system of two objects, it is possible to have a momentum of zero. One example would be two carts released from an internal spring “explosion”. The total momentum is zero, but both of the individual parts have kinetic energy. If the system of objects has momentum, then at least one of them is moving, and there will be kinetic energy. It is not possible to have momentum and zero kinetic energy at the same time.
10. m1=0.15 kg 2 1 0.15 kg 22 m/s [N] m v = =
Choose north as positive. 1 1 2 2 1 1 2 2 1 1 2 2 1 2 1 1 2 2 1 2 ( ) (0.15 kg)(22 m/s) (0.15 kg)( 22 m/s) 0.15 kg 0.15 m/s 0 m/s m v m v m v m v m v m v m m v m v m v v m m v ′ ′ + = + ′ + = + + ′ = + + − = + ′ =
The velocity of the two-ball system after the collision is 0 m/s.
11. Conservation of Momentum 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 1
(but 0 and the masses are equal) 0 (Equation 1) m v m v m v m v v mv mv mv v v v v v v ′ ′ + = + = ′ ′ + = + ′ ′ = + ′ = − ′ Conservation of Energy 2 2 2 2 1 1 1 1 1 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 1 1 2 2 2 2 2 1 1
(but 0 and the masses are equal) 0 (Equation 2) m v m v m v m v v mv mv mv v v v v v v ′ ′ + = + = ′ ′ + = + ′ ′ = + ′ = − ′ Substitute Equation 1 into Equation 2:
2 2 2 1 1 1 1 2 2 2 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 ( ) 2 2 2 0 ( ) 0 0 or 0 v v v v v v v v v v v v v v v v v v v ′ ′ − = − ′ ′ ′ − + = − ′ ′ / −/ = ′ ′ − = ′= ′− =
If v1′− =v1 0, then v1′=v1 (i.e., the speed of the proton after the collision is unchanged). This is not possible, so v1′=0. The final speed of the first proton is zero after the collision.
Substituting back into Equation 1 gives:
2 1 1 1 1 2 0 815 m/s v v v v v v ′= − ′ = − = ′ =
The final velocity of the second proton after the collision is 815 m/s in the direction of the initial velocity. 12. For an inelastic collision, the vehicles stick together.
4 1 3 2 1 2 1 2 1.3 10 kg 1.1 10 kg 90 km/h [N] 30 km/h [N] ? (when coupled) m m v v v v v = × = × = = ′= ′ = =′
Choose north as positive. 1 1 2 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 2 4 3 4 3 when coupled, ( ) (1.3 10 kg)(90 km/h) (1.1 10 kg)(30 km/h) 1.3 10 kg 1.1 10 kg 85 km/h [N] m v m v m v m v v v v m v m v m m v m v m v v m m v ′ ′ + = + ′= ′ = ′ ′ + = + + ′ = + × + × = × + × ′ =
The velocity of the vehicles is 85 km/h [N] just after the collision.
13. 4 t 1.3 10 kg m = × 3 c t c 1.1 10 kg 90 km/h [N] 25 m/s [N] 30 km/h [N] 8.333 m/s [N] m v v = × = = = = T K truck K car 2 2 t t c c 4 2 3 2 6 T 1 1 2 2 1 1 (1.3 10 kg)(25 m/s) (1.1 10 kg)(8.333 m/s) 2 2 4.1 10 J E E E m v m v E = + = + = × + × = ×
The final velocity of the vehicles after the collision is 85.319 km/h [N] = 23.7 m/s [N].
T K truck K car 2 2 t t c c 4 2 3 2 6 T 1 1 2 2 1 1 (1.3 10 kg)(23.7 m/s) (1.1 10 kg)(23.7 m/s) 2 2 4.0 10 J E E E m v m v E ′ = ′ + ′ ′ ′ = + = × + × ′ = ×
The decrease in kinetic energy is 4.1 × 106 – 4.1 × 106 = 1 × 105 J. 14. Choose the original direction of motion as the positive direction.
26 O 26 N O 2 O N N 5.31 10 kg 4.65 10 kg 0 4.81 10 m/s 34.1 m/s ? m m v v v v − − = × = × = ′ = × ′ = − = N N O O N N O O N N N N O O N N O O N N O O N N 26 2 0 (5.31 10 kg)(4.81 10 m/s) 34.1 m/s m v m v m v m v m v m v m v m v m v v m m v v m − ′ ′ + = + ′ ′ + = + ′ + ′ = ′ ′ = + × × = − +
Applying Inquiry Skills
15. (a) Line A represents the total momentum because the total momentum in the system is constant.
(b) The collision is an inelastic one. If the collision were elastic, the total kinetic energy before and after would have been the same.
Making Connections
16. (a) The rubber bullet would have the elastic collision and the lead bullet would have the inelastic collision. (b) ∆ = −∆pGT pGR m vT T2−m vT T1= −(m vR R2−m vR R1) G G G G vT1=0, and vR2= −vR1 G G G T T2 R R1 R R1 T T2 R R2 R R1 R1 R2 0 ( ) m v m v m v m v m v m v v v − = − − = + = G G G G G G G G T T2 R R2 R R1 T R R m v m v m v p p p = + ′ = ′ + G G G G G G T L T T2 T T1 L L2 L L1 T1 0 p p m v m v m v m v v ∆ = −∆ − = − + = G G G G G G G T T2 L L2 L L2 T T2 L L1 L L1 T L L 0 m v m v m v m v m v m v p p p − = − = − ′ = − ′ G G G G G G G G G
The rubber bullet transfers more momentum to the target.
(c) Rubbers bullets are preferred in crowd control because they are less likely to kill or permanently injure any of the crowd and they impart a larger backward impulse on the crowd.
Section 5.3 Questions
(Page 253)
Understanding Concepts
1. (a) It is not possible for both objects to be at rest. If they were both at rest, the initial momentum of the first object would have violated the law of conservation of t momentum.
(b) It is possible for the first object to be at rest after the collision. One example is a curling stone that strikes another and then stops moving.
2. The does not violate the law of conservation of momentum for the system which contains the earth and the snowball. The earth exerts a net external force on the tree/snowball system. This external force negates the conservation of momentum for that system.
3. There momentums will only be the same if they have the same mass. The relationship between momentum and kinetic energy is 2 K 2 2 2 2 K K 2 2 ( ) 2 2 2 mv E m v m mv m p E m p mE = = = = =
4. Conservation of Momentum 1 1 2 2 1 1 2 2 2 1 1 1 1 2 2 1 1 1 1 2 2 1 1 1 2 2 (but 0) 0 ( ) (Equation 1) m v m v m v m v v m v m v m v m v m v m v m v v m v ′ ′ + = + = ′ ′ + = + ′ ′ − = ′ ′ − = Conservation of Energy 2 2 2 2 1 1 1 1 1 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 1 1 1 1 2 2 2 2 2 1 1 1 2 2 (but 0) 0 ( ) (Equation 2) m v m v m v m v v m v m v m v m v m v m v m v v m v ′ ′ + = + = ′ ′ + = + ′ ′ − = ′ ′ − =
Divide Equation 2 by Equation 1:
2 2 2 1 1 1 2 2 1 1 1 2 2 1 1 1 1 2 1 1 2 1 1 1 2 1 ( ) ( ) ( )( ) ( ) (Equation 3) or (Equation 4) m v v m v m v v m v v v v v v v v v v v v v v ′ ′ − = ′ ′ − ′ ′ + − = ′ ′ − ′ = + ′ ′= −′
Substitute Equations 3 and 4 back into the original conservation of momentum equation:
1 1 1 1 2 2 1 1 1 1 2 1 1 1 1 1 1 2 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 2 1 1 2 1 1 1 2 1 0 ( ) ( ) ( ) 0.022 kg 0.027 kg (3.5 m/s) 0.022 kg 0.027 kg 0.36 m/s [forward], or m v m v m v m v m v m v v m v m v m v m v m v m v m v m v v m m m m v m m v v m m v ′ ′ + = + ′ ′ = + + ′ ′ = + + ′+ ′= − ′ + = − − ′ = + − = + ′ = − 0.36 m/s [backward]
The velocity of the 22-g superball after the collision is 0.36 m/s [backward] .
1 1 1 1 2 2 1 1 1 2 1 2 2 1 1 1 2 1 1 2 2 1 2 2 2 1 1 2 1 2 1 1 1 2 1 1 2 2 0 ( ) 2 ( ) 2 2 2(0.022 kg) (3.5 m/s) 0.022 kg 0.027 kg 3.1 m/s [forward] m v m v m v m v m v v m v m v m v m v m v m v m v m v v m m m v m v v m m v ′ ′ + = + ′ ′ = − + ′ ′ = − + ′+ ′= ′ + = ′ = + = + ′ =
5. Using the equations derived in question 4 , 1 2 1 1 1 2 2 1 1 2 2 2 2 2 2 2 1 3 3( ) 3 3 4 2 1 2 m m v v m m m m v v m m m m m m m m m m m m m m − ′ = + − = + + = − + = − = = 6. m=66 kg 2 1 9.8 m/s 25 m ? g y v = ∆ = =
Use conservation of energy to solve for the speed of the moving skier at the bottom of the hill.
T T 2 1 1 2 1 1 2 2 2(9.8 m/s )(25 m) 22.14 m/s E E mg y mv v g y v ′ = ∆ = = ∆ = = 2 1 2 1 2 72 kg 22.14 m/s 0 ? (when coupled) m v v v v v = = = ′= ′ = =′ 1 1 2 2 1 1 2 2 2 1 2 1 1 1 2 1 1 1 2
0, and when coupled,
0 ( ) (66 kg)(22.14 m/s) 66 kg 72 kg 11 m/s m v m v m v m v v v v v m v m m v m v v m m v ′ ′ + = + ′ ′ ′ = = = ′ + = + ′ = + = + ′ =
The speed of the two-skier system immediately after the collision is 11 m/s.
Applying Inquiry Skills
7. (a) The collision takes place very quickly. The duration of time that the bullet and block are interacting, the strings are vertical, and only balance the gravitational forces. During the collision (not during the swing), momentum is conserved. (b) Let V be the speed of the bullet and block combination after the collision.
1 1 2 2 1 1 2 2 (but 2 0 and when coupled, 1 2 )
0 ( ) m v m v m v m v v v v V mv m M V mv V m M ′ ′ ′ ′ + = + = = = + = + = +
(d) ET =ET′ 2 2 2 2 2 2 2 1 2 1 2 2 2 2 ( ) mg y mv gh V V h g mv m M g m v h g m M ∆ = = = + = = + (e) 2 2 2 2 ( ) m v h g m M = + 2 2 2 2 2 2 2 ( ) 2 ( ) 2 2 m v gh m M gh m M v m m M gh m m M v gh m = + + = + = + = (f) m=87 g 0.0087 = kg 2 5.212 kg 9.8 m/s 6.2 cm 0.062 m ? M g h v = = = = = 2 2 2 0.0087 kg 5.212 kg 2(9.8 m/s )(0.062 m) 0.0087 kg 6.6 10 m/s m M v gh m v + = + = = ×
The initial speed of the bullet was 6.6 10 m/s× 2 .
(g) Some of the sources of error would be friction of the moving parts of the gun and loss of energy to thermal energy in the spring. The frictional draw of the catch mechanism would use some of the energy that should be converted into gravitational potential.
8. There are a number of advantages of a crumple zone. One is that it converts a significant part of the kinetic energy of the vehicle into thermal energy as the steel become permanently deformed. The crumple zone also causes the vehicle to slow down over a greater distance. This greater distance increases the length of time the vehicle is slowing down, and the average force to stop the vehicle (and its passengers) decreases.
9. Most meteorites burn up in the atmosphere. Larges known one is about 60 metric tons, and the next known one is about 30 metric tons. Large collisions have rarely (ever 300 million years or so) and can cause devastating climactic change.
