Question

A block of mass mi 1.60 kg moving at v1 2.00 m/s undergoes a completely inelastic collision with a Part A telLionary block of mass m = 0.600 kg. The blocks time, the two-block system collides inelastically with a third block, of mass m3 = 2.50 kg , which is initially at rest. The three blocks then move, stuck together, with speed u3.(Figure 1) Assume that the blocks slide without friction. Find the ratio of the velocity v2 of the two-block system after the first collision to the velocity vi of the block of mass m before the collision. Express your answer numerically using three significant figures. View Available Hint(s) Figure 1 of 1 Πνα ΑΣφ ? Previous Answers Submit 2 m7 Incorrect; Try Again; 4 attempts remaining 3 Part B my the ratio of the velocity v of the three-block system after the second collision to the velocity v1 of the block Find mass mi before the collisions.

Answer 1

A)

Using law of conservation of momentum,

V2/v1 = m1/(m1 + m2) = 1.6/(1.6 + 0.6) = 0.727

​B)

Using law of conservation of momentum,

V3/v1 = 1.6/(1.6 + 0.6 + 2.5) = 0.34

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