Block 1, of mass m 1 = 9.10kg , moves along a frictionless air track with speed

Question

Block 1, of mass m1 = 9.10kg , moves along a frictionless air track with speed v1 = 29.0m/s . It collides with block 2, of mass m2 = 13.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

Figure 1 of 1

Part A

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

Express your answer numerically.

Part B

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

Express your answer numerically.

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.

pi =   kg?m/s   Block 1, of mass m1 = 9.10kg , moves along a frictionless air track with speed v1 = 29.0m/s . It collides with block 2, of mass m2 = 13.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) Figure 1 of 1 Part A Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. pi = kg?m/s Part B Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. vf = m/s 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. ?K = J

Explanation / Answer

A> initial momentum = 9.10 x 29 + 13 x 0 = 263.9 kg.m/s

B>

using momentum conservation during collision,

9.10 x 29 + 13 x 0 = (9.10 +13)v

v =11.94 m/s

c) change in K = kfinal - Kinitial

= (9.10 + 13) x 11.94^2 /2 - 9.10 x 29^2 /2 - 0 = - 2251.22 J

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