Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed
ID: 1613808 • Letter: B
Question
Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed v1 = 17.0 m/s . It collides with block 2, of mass m2 = 51.0 kg , 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=KfinalKinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules. if you can show your work just so I can follow along, and correct my mistakes
Explanation / Answer
m1 = 2.7 kg , u1 = 17 m/s, m2 = 51 kg , u2 = 0
(a) pi = m1u1 +m2u2 = 2.7*17
pi = 45.9 kg.m/s
(b)from conservation of momentum
pi = pf = (m1+m2)vf
45.9 = (2.7 +51)vf
vf =0.855 m/s
(c) K=KfinalKinitial
= (0.5*(m1+m2)vf^2) - (0.5*m1u1^2) -0
= (0.5*(2.7+51)*0.855^2) - (0.5*2.7*17*17)
= -370.5 J