Physics Question?

Question

Block 1, of mass "m1", moves across a frictionless surface with speed "vi". It collides elastically with block 2, of mass "m2", which is at rest (vi=0). After the collision, block 1 moves with speed "vf", while block 2 moves with speed "vf". Assume that m1 > m2, so that after the collision, the two objects move off in the direction of the first object before the collision.

1. This collision is elastic. What quantities, if any, are conserved in this collision?
-kinetic energy only
-momentum only
-kinetic energy and momentum

2. What is the final speed vf of block 1?
Express vf in terms of m1, m2, and vi.

3. What is the final speed vf of block 2?
Express vf in terms of m1, m2, and vi.

Answer 1

This one is straight out of the physics book, equations and all.

1. It's an elastic collision, so both kinetic energy and momentum are conserved.

2. vf = vi (m1 - m2) / (m1 + m2) (Eq. 9.22, p. 175, physics)

3. vf = vi (2m1) / (m1 + m2) (Eq. 9.23, p. 176, physics)

Answer 2

1. In all collisions, momentum is conserved. In elastic collisions, kinetic energy is also conserved.

2. Momentum before collision: m1*v1
Kinetic energy before collision = .5*m1*v1^2

Momentum after collision m1*v1* + m2*v2* where v1* and v2* are the after-collision velocities of the masses.

Conserve momentum: m1*v1 = m1*v1* + m2*v2*; you can solve for v1*, but you do not know v2*. Use energy conservation then:

Kinetic energy after collision = .5*m1*v1*^2 + .5*m2*v2*^2; conserving energy gives the relation

.5*m1*v1^2 = .5*m1*v1*^2 + .5*m2*v2*^2.

Solve for v2* in terms of m1, m2, v1 and v1*, substitute for v2* in the momentum equation which will then contain only m1, m2, v1 and v1*. Solve it for v1* to get your answer.

3. Once you get v2* you can use either energy conservation or momentum conservation to get v2*