Three-Block Inelastic Collision A block of mass m1 = 1.90 kg moving at v1 = 1.10

Question

Three-Block Inelastic Collision
A block of mass m1 = 1.90 kg moving at v1 = 1.10 m/s undergoes a completely inelastic collision with a stationary block of mass m2= 0.400kg . The blocks then move, stuck together, at speed v2 . After a short time, the two-block system collides inelastically with a third block, of mass3 = 2.70kg , which is initially at rest. The three blocks then move, stuck together, with speed v3. Assume that the blocks slide without friction.

A. 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.
v2/v1=
B.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
v3/v1=

Explanation / Answer

Given that m1 = 1.90 kg V1 = 1.1 m/s m2 = 0.4 kg and is initially rest. The combined speed, V2 = m1V1 / (m1 + m2) = 0.908 m/s Now the mass of two block system, M = m1 + m2 = 2.3 kg The speed of two block system, V2 = 0.908 m/s Mass of third block, m3 = 2.70 kg which is initially rest. Then the final velocity after collision, V3 = MV2 / (M + m3) = 0.65 m/s A) V2/V1 = 0.908 / 1.1 = 0.83 B) V3/V1 = 0.65/1.1 = 0.59

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