A block of mass1 = 2.00kg moving at v1= 2.00m/s undergoes a completely inelastic
ID: 2178050 • Letter: A
Question
A block of mass1 = 2.00kg moving at v1= 2.00m/s undergoes a completely inelastic collision with a stationary block of mass2 = 0.800 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 3= 2.70 , which is initially at rest. The three blocks then move, stuck together, with speed v3.(Figure 1) Assume that the blocks slide without friction.
a) Find v2/v1, the ratio of the velocity of the two-block system after the first collision to the velocity v1 of the block of mass m1 before the collision.
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.
A block of mass1 = 2.00kg moving at v1= 2.00m/s undergoes a completely inelastic collision with a stationary block of mass2 = 0.800 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 3= 2.70 , which is initially at rest. The three blocks then move, stuck together, with speed v3.(Figure 1) Assume that the blocks slide without friction. a) Find v2/v1, the ratio of the velocity of the two-block system after the first collision to the velocity v1 of the block of mass m1 before the collision. 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.Explanation / Answer
V2 = 2* 2 / 2.8 = 1.428 m/s V3 = 1.428 * 2.8/ 5.5 = 0.72 m/s a) V2 / V1 = 1.428 / 2 = 0.714 m/s b)V3 / V1 = 0.72 / 2 = 0.36 m/s
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