A block of mass m1 = 1.40kg moving at v1 = 2.00m/s undergoes a completely inelas

Question

A block of mass m1 = 1.40kg moving at v1 = 2.00m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.900kg . 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 = 2.20kg , 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.

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 answers numerically using three significant figures.

Explanation / Answer

in inelastic collison,

intial momentum = final momentum,

m1v1=(m1+m2)v2,

v2= m1v1/(m1+m2)

m1v1= (m1+m2+m3)v3,

v3 = m1v1/(m1+m2+m3)

just put the values and u;'ll get the right answer,enjoy :)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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