A block of mass m1 = 1.70 kg moving at U-1.40 m/s undergoes a completely inelast

Question

A block of mass m1 = 1.70 kg moving at U-1.40 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.900 kg . The blocks then move, stuck together, at speed U2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 3.00 kg , which is initially at rest. The three blocks then move, stuck together, with speed 3.(Figure 1) Assume that the blocks slide without friction. 02 Find, the ratio of the velocity v2 of the two-block system after the first collision to the velocity ul of the block of mass mi before the collision Express your answer numerically using three significant figures Hints U2 Submit My Answers Give Up Part B Find th ratio of the velocity s of the three-block system after the second collision to the velocity "1 of the block of mass m before the collisions Figure 1 of 1 Express your answer numerically using three significant figures Hints Submit My Answers Give Up /n 1 Im Continue

Explanation / Answer

by conservation of momentum

initial momentum = final momentum

m1 * v1 = (m1 + m2) * v2

v2 / v1 = m1 / (m1 + m2)

v2 / v1 = 1.7 / (1.7 + 0.9)

A) v2 / v1 = 0.6538

v2 / 1.4 = 0.6538

v2 = 0.915 m/s

again by conservation of momentum

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

(1.7 + 0.9) * 0.915 = (1.7 + 0.9 + 3) * v3

v3 = 0.4248 m/s

v3 / v1 = 0.4248 / 1.4

B) v3 / v1 = 0.3034

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