Block 1, of mass m 1 = 1.50 k g , moves along a frictionless air track with spee
ID: 2143738 • Letter: B
Question
Block 1, of mass m1 = 1.50kg , moves along a frictionless air track with speed v1 = 29.0m/s . It collides with block 2, of mass m2 = 27.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)
Find the magnitude pi of the total initial momentum of the two-block system.
Block 1, of mass m1 = 1.50kg , moves along a frictionless air track with speed v1 = 29.0m/s . It collides with block 2, of mass m2 = 27.0kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) Find the magnitude pi of the total initial momentum of the two-block system. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. 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
Pi = m1*vi = 43.5 kg.m/s
m1*vi = (m1+m2)*vf
vf = m1*vi/(m1+m2)
vf = 1.526 m/s
KEi = 0.5*m1*vi^2 = 630.75 J
KEf = 0.5*(m1+m2)*vf^2 = 33.18 J
delta K = KEf - KEi = -597.6 J