Block 1, of mass m1 = 5.50kg , moves along a frictionless air track with speed v

Question

Block 1, of mass m1 = 5.50kg , moves along a frictionless air track with speed v1 = 17.0m/s . It collides with block 2, of mass m2 = 11.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

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

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

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

Block 1, of mass m1 = 5.50kg , moves along a frictionless air track with speed v1 = 17.0m/s . It collides with block 2, of mass m2 = 11.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) A) Find the magnitude pi of the total initial momentum of the two-block system. B) Find vf, the magnitude of the final velocity of the two-block system C) What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision?

Explanation / Answer

Part A)

Initial momentum = (mv)

Initial momentum = (5.5)(17) = 93.5 kg m/s

Part B)

Initial momentum = Final Momentum

93.5 = (M+m)v

93.5 = (5.5 + 11)(v)

v = 5.67 m/s

Part C)

change in KE = .5(16.5)(5.67)2 - .5(5.5)(17)2

change in KE = -529 J (Negative since it loses Energy)

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