A block of mass m1 = 1.40 kg moving at v1 = 1.00 m/s undergoes a completely inel

Question

A block of mass m1 = 1.40 kg moving at v1 = 1.00 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.500 kg . The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 2.70 kg , which is initially at rest. The three blocks then move, stuck together, with speed v3.(Figure 1) Assume that the blocks slide without friction.

Find v2v1, 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.

Find v3v1, 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.

Explanation / Answer

During a collision momentum of system of two masses remains conserved. Writing expression of momentum conservation for first collision, we get

M1 V1 = (M1 + M2) V2   Hence V2/V1  = M1 /( M1 + M2) = 0.737

similarly for seccond collision we get

(M1 + M2 ) V2 = ( M1 + M2 + M3 ) V3 = M1 V1   ( last term from above equation )

Hence V3 / V1 = M1 / ( M1 + M2 + M3 ) = 0.304

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