A block of mass m_1 = 1.40kg moving at v_1 = 1.80m/s undergoes a completely inel

Question

A block of mass m_1 = 1.40kg moving at v_1 = 1.80m/s undergoes a completely inelastic collision with a stationary block of mass m_2 = 0.800kg . 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 = 2.70kg , 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. Part A Find rac{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. rac{v_2}{v_1} = 0.636 Part B Find rac{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

m1 = 1.4 kg m2 = 0.8 kg m3 = 2.7 kg v1 = 1.8 m/s by conservation of momentum after collision m1v1 = (m1+m2)v2 => v2 = 1.145 m/s similarily applying conservation of momentum in second collision (m1+m2)v2 = (m1+m2+m3)v3 => v3 = 0.5141 m/s Ans(a)=>v2/v1 = 0.636 ans(b)=>v3/v1 = 0.286

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