Block 1, of mass m1 = 5.70kg , moves along a frictionless air track with speed v

Question

Block 1, of mass m1 = 5.70kg , moves along a frictionless air track with speed v1 = 21.0m/s . It collides with block 2, of mass m2 = 53.0kg , which was initially at rest. The blocks stick together after the collision

Part A

Find the magnitude pi of the total initial momentum of the two-block system.

Express your answer numerically.

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Part B

Find vf, the magnitude of the final velocity of the two-block system.

Express your answer numerically.

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Part C

What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision?

Express your answer numerically in joules.

Explanation / Answer

A. initial momentum Pi = 5.70 x 21 + 53 x 0 = 119.7 kg.m/s

B. from momentum conservation

initial momentum = final momentum

119.7 = (5.7 + 53) v

v = 2.04 m/s

C. delta K = final K - initial K

= ( 5.7 x 21^2 /2 ) - ( (5.7+53)x 2.04^2 /2 )

= 1134.71 J

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