Question

A block of mass m_1 = 1.80 kg moving at v_1 = 2.00 m/s undergoes a completely inelastic collision with a stationary block of mass m_2 = 0.300 kg. 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 = 2.20 kg, which is initially at rest. The three blocks then move, stuck together, with speed v_3. (Figure 1) 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. Express your answer numerically using three significant figures. 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. Express your answer numerically using three significant figures.

Answer 1

momentum conservation suggests that

$\frac{v_{2}}{v_{1}}= \frac{m_{1}}{m_{1}+m_{2}} = 0.857$

$\frac{v_{3}}{v_{1}}= \frac{m_{1}}{m_{1}+m_{2}+m_{3}} = 0.419$