Block 1, of mass m1 = 8.10kg , moves along a frictionless air track with speed v

Question

Block 1, of mass m1 = 8.10kg , moves along a frictionless air track with speed v1 = 13.0m/s . It collides with block 2, of mass m2 = 55.0kg , 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.

Part B:

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

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

IMAGE

Block 1, of mass m1 = 8.10kg , moves along a frictionless air track with speed v1 = 13.0m/s . It collides with block 2, of mass m2 = 55.0kg , 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. Part B: Find vf, the magnitude of the final velocity of the two-block system. Part C:What is the change ?K=Kfinal?Kinitial in the two-block system's kinetic energy due to the collision? IMAGE

Explanation / Answer

conservation of momentum

a) magnitude of initial momentum = m1v1+m2v2

= 8.1 *13 + 55 * 0

= 105.3 kg m

b)conservation of momentum

  m1v1+m2v2 = (m1+m2)vf

vf = final velocity of the system

8.1 *13 + 55 * 0 = (8.1+55)vf

final velocity ,vf = 1.669 m/s

c) initial KE , ki = 1/2 m1v12 + 1/2 m2v22

= 1/2 *8.1 * 132 + 0

= 684.45 J

final KE , kf = 1/2 (m1+m2)*vf2

= 1/2 (8.1+55) *1.662

= 86.94 J

change in KE = kf - ki

=684.45 - 86.94

= 597.51 J

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