A block of mass = 1.70 moving at = 1.60 undergoes a completely inelastic collisi

Question

A block of mass = 1.70 moving at = 1.60 undergoes a completely inelastic collision with a stationary block of mass = 0.500 . The blocks then move, stuck together, at speed . After a short time, the two-block system collides inelastically with a third block, of mass = 3.00 , which is initially at rest. The three blocks then move, stuck together, with speed . Assume that the blocks slide without friction. Part A Find v2/v1, the ratio of the velocity v2 of the two-block system after the first collision to the velocity v1 of the block of mass m1 before the collision. Part B Find m3/m1, the ratio of the velocity of the three-block system after the second collision to the velocity v1 of the block of mass m1 before the collisions. =

Explanation / Answer

P1 = P2
m1*v1 = (m1+m2)*v2
v2/v1 = m1/(m1+m2) = 1.7/(1.7+.500) = .777


P1 = P3
m1*v1 = (m1+m2+m3)*v3
v3/v1 = m1/(m1+m2+m3) = .326

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