A block of mass m1 = 2.00 kg moving at v1 = 1.60 m/s undergoes a completely inel

Question

A block of mass m1 = 2.00 kg moving at v1 = 1.60 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.600 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. Assume that the blocks slide without friction. Part A Find v2/v1, 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.answer must be in terms of v2/v1= Part 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.Ansewr must be in terms of v3/v1=

Explanation / Answer

Part A:
for momentum conservation, after the first collision, we have:
m1 *v1 = (m1+m2)*v2

v2 / v1 = m1 / (m1+m2) = 2.00kg/(2.00kg+0.600kg) = 0.769 (1)

PART B

from (1), we have:

v2 = m1 v1 / (m1+m2) (2)

for momentum conservation, after the second collision, we have:

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

using (2), we have:

(m1+m2)*[ m1 v1 / (m1+m2)] = (m1+m2 +m3)*v3

=> m1 v1 = (m1+m2 +m3)*v3

v3 / v1 = m1 / (m1+m2 +m3) = 2.00kg / (2.00kg+0.600kg + 2.50kg) = 0.392

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.