A block of mass m_2 moving with speed v_1 undergoes a completely inelastic colli

Question

A block of mass m_2 moving with speed v_1 undergoes a completely inelastic collision with a stationary block of mass m_2. The blocks then move, stuck together, at speed v_2. After a short time, the two-block system collides inelastically with a third block, of mass m_3, which is initially stationary. The three blocks then move, stuck together, with speed (Figure 1) . All three blocks have nonzero mass. Assume that the blocks slide without friction.

Find v_2/v_1, the ratio of the velocity v_2 of the two-block system after the first collision to the velocity v_1 of the block of mass m_1 before the collision??

Find K_2/K_1, the ratio of the kinetic energy K_2 of the two-block system after the first collision to the kinetic energy K_1 of the block of mass m_1 before the collision.

Find v_3/v_1, the ratio of the velocity v_3 of the three-block system after the second collision to the velocity v_1 of the block of mass m_1 before the collisions.


Find K_3/K_1, the ratio of the kinetic energy K_3 of the three-block system after the second collision to the initial kinetic energy K_1 of the block of mass m_1 before the collisions

Suppose a fourth block, of mass m_4, is included in the series, so that the three-block system with speed v_3 collides with the fourth, stationary, block. Find K_4/K_1, the ratio of the kinetic energy K_4 of all the blocks after the final collision to the initial kinetic energy K_1 of the block of mass m_1 before any of the collisions.
Express your answer in terms of m_1, m_2, m_3, and/or m_4.

Explanation / Answer

in inelastic collision, m1v1 = (m1+m2)v2

v2/v1 = m1/(m1+m2)

K2/K1 = [(m1+m2)v2^2 ]/[m1v1^2]

v3/v1 = m1/[m1+ m2 + m3]

K3/K1 = [(m1+ m2 + m3)v3^2]/[m1v1^2]

v4/v1 = m1/[m1+ m2 + m3 +m4]

K4/K1 = [(m1+ m2 + m3 +m4)v4^2]/[m1v1^2]

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