A block of mass m1 = 1.80kg moving at v1 = 1.40 m/s undergoes a completely inela

Question

A block of mass m1 = 1.80kg moving at v1 = 1.40 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.700kg The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 2.10kg which is initially at rest. The three blocks then move, stuck together, with speed v3. (Figure 1) Assume that the blocks slide i without friction. 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. Express your answer numerically using three significant figures. Find v3/v1, the ratio of the velocity v3 of the three-block system after the second collision to the velocity v1 of the block of mass m1 before the collisions. Express your answer numerically using three significant figures.

Explanation / Answer

P1 = P2 m1*v1 = (m1+m2)*v2 ? v2/v1 = m1/(m1+m2) = 1.1/(1.4+.700) = .524 P1 = P3 m1*v1 = (m1+m2+m3)*v3 ? v3/v1 = m1/(m1+m2+m3) = .25

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