Three-Block Inelastic Collision A block of mass m 1 = 1.90 kg moving at v 1 = 1.
ID: 1461387 • Letter: T
Question
Three-Block Inelastic Collision
A block of mass m1 = 1.90 kg moving at v1 = 1.60 m/sundergoes a completely inelastic collision with a stationary block of mass m2 = 0.700 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 = 2.50 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.
Part A
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.
Express your answer numerically using three significant figures.
Part B
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.
Express your answer numerically using three significant figures.
Figure 1 of 1
Three-Block Inelastic Collision
A block of mass m1 = 1.90 kg moving at v1 = 1.60 m/sundergoes a completely inelastic collision with a stationary block of mass m2 = 0.700 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 = 2.50 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.
Part A
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.
Express your answer numerically using three significant figures.
v2v1 =Part B
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.
Express your answer numerically using three significant figures.
v3v1 =Figure 1 of 1
Explanation / Answer
A) Apply conservation of momentum
m1*v1 = (m1+m2)*v2
v2/v1 = m1/(m1+m2)
= 1.9/(1.9 + 0.7)
= 0.73 <<<<<-----------Answer
B) v2 = 0.73*v1
= 0.73*1.6
= 1.17 m/s
Agaian aplly conservation of momentum at second collision.
(m1+m2)*v2 = (m1+m2+m3)*v3
==> v3/v2 = (m1+m2)/(m1+m2+m3)
= (1.9 + 0.7)/(1.9 + 0.7 + 2.5)
= 0.51
v3/(0.73*v1) = 0.51
v3/v1 = 0.51*0.73
v3/v1 = 0.372 <<<<<-----------Answer
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