A block of mass m_1 = 1.40 kg moving at v_1= 2.00 m/s undergoes a completely ine

Question

A block of mass m_1 = 1.40 kg moving at v_1= 2.00 m/s undergoes a completely inelastic collision with a stationary block of mass m_2 = 0.700 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 = 2.50 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.

Explanation / Answer

Apply conservation of momentum after the first collision,
m1 v1 = (m1+m2)v2

v2 / v1 = m1 / (m1+m2)

= 1.4kg/(1.4kg+0.7kg)

= 0.666

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from the first equation

v2 = m1 v1 / (m1+m2)

for momentum conservation, after the second collision,

(m1+m2)*v2 = (m1+m2 +m3)*v3

(m1+m2)*[ m1v1/ (m1+m2)] = (m1+m2+m3)*v3

m1 v1 = (m1+m2 +m3)*v3

v3 / v1 = m1 / (m1+m2 +m3)

=1.4kg/(1.4kg+0.7kg+2.5 kg)

= 0.304

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