Blocks A (mass 7.00 kg ) and B (mass 12.00 kg , to the right of A) move on a fri

Question

Blocks A (mass 7.00 kg ) and B (mass 12.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A.

Part A

Find the maximum energy stored in the spring bumpers.

Express your answer with the appropriate units.

Uspringmax =

Part B

Find the velocity of block A when the energy stored in the spring bumpers is maximum.

Express your answer with the appropriate units.

vA =

Part C

Find the velocity of block B when the energy stored in the spring bumpers is maximum.

Express your answer with the appropriate units.

vB=

Part D

Find the velocity of block A after the blocks have moved apart.

Express your answer with the appropriate units.

vA=

Part E

Find the velocity of block B after the blocks have moved apart.

Express your answer with the appropriate units.

vB=

Explanation / Answer

Part A The maximum spring compression occurs when the velocities are the same (otherwise the spring is compressing or decompressing). To find the common velocity, conserve momentum:
7kg * 2m/s - 12kg*0.5m/s = 19kg * v
v = 0.42 m/s

initial KE = ½ * 7 * (2)² = 14 J
final KE = ½ * 19* (0.42)² = 1.676 J
so max U = 12.324J

Part D

Begin again with conservation of momentum:
7kg * 2m/s-12kg*0.5m/s = 7kg * u + 12kg * v
for u, v the final velocity of A and B, respectively.
8 =7u + 12v

For an elastic, head-on collision, we know (from CoE) that
the relative velocity of approach = relative velocity of separation, or
2+0.5 = v - u, so
v = u + 2.5m/s plug into momentum equation
u=-1.157m/s ,

Part E v=1.343 m/s

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