Blocks A (mass 7.00 kg ) and B (mass11.00 kg , to the right of A ) move on a fri

Question

Blocks A (mass 7.00 kg ) and B (mass11.00 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.

A)Find the maximum energy stored in the spring bumpers.

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

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

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

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

Explanation / Answer

We know that the maximum spring compression occurs when the velocities are the same. So we need to find the common velocity, conserve momentum:

7kg * 2m/s + 11*0.5 = 18kg * v

v = 1.08 m/s

initial KE =0.5 *7kg * (2m/s)2 = 14 J

initial KE =0.5 *11kg * (0.5m/s)2 = 1.375 J

final KE = 0.5* 18kg * (1.08m/s)2 = 10.49 J

so max energy stored is,U = 4.885 J

b)maximum velocity will be

7kg * 2m/s + 11*0.5 = 18kg * v

v = 1.08 m/s

1.08 m/s

c)maximum velocity will be

7kg * 2m/s + 11*0.5 = 18kg * v

v = 1.08 m/s

1.08 m/s

D)consider conservation of momentum:

7kg * 2m/s = 7kg * u + 11kg * v

for u, v the final velocity of A and B, respectively.

Dividing by 7kg gives us

2m/s = u + 1.57v

For an elastic, head-on collision, we know (from CoE) that the relative velocity of approach = relative velocity of separation, or

2 m/s = v - u, so

v = u + 2m/s put it into momentum equation

2m/s = u + 1.57(u + 2m/s) = 2.57u + 3.14m/s

u =- 0.44m/s

E) v = u + 2m/s = 1.555m/s

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