Block 1, of mass m1 = 9.50 kg , moves along a frictionless air track with speed
ID: 1538171 • Letter: B
Question
Block 1, of mass m1 = 9.50 kg , moves along a frictionless air track with speed v1 = 13.0 m/s . It collides with block 2, of mass m2 = 41.0 kg , which was initially at rest. The blocks stick together after the collision
1. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.
2.Find vf, the magnitude of the final velocity of the two-block system. Express the answer numerically.
3.What is the change K=KfinalKinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.
Explanation / Answer
1) initial total momentum = m1v1+m2v2 = 9.5*13+0 = 123.5 kg m/s
b) from ci=onservation of momentum
m1v1+m2v2 = (m1+m2)v
v = 123.5/50.5
v = 2.45 m/s
3) kf-ki = 1/2(50.5)2.45^2-1/2*9.5*13^2 = 151.56-802.75 = -651.19 J