Block 1, of mass m1 = 9.30kg , moves along a frictionless air track with speed v

Question

Block 1, of mass m1 = 9.30kg , moves along a frictionless air track with speed v1 = 23.0m/s . It collides with block 2, of mass m2 = 51.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 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 = 9.30kg , moves along a frictionless air track with speed v1 = 23.0m/s . It collides with block 2, of mass m2 = 51.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 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.

Explanation / Answer

A) Pi = m1*v1 + m2*v2

= 9.3*23 + 0

= 213.9 kg.m/s

B) Pi = Pf

Pi = (m1+m2)*vf

vf = Pi/(m1+m2)

= 213.9/(9.3+51)

= 3.55 m/s

C) delta k = Kf - Ki

= 0.5*(m1+m2)*vf^2 - 0.5*m1*v1^2

= 0.5*(9.3+51)*3.55^2 - 0.5*9.3*23^2

= -2080.5 J

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