I need some help, I can't solve this physics problem.

Question

Answer 1

First ... use conservation of energy (mgh=.5mv²) to find the velocity of m₁ just before the collision with m₂ to be 0.713 m/s.

Second use conservation of momentum (.5*.713=.5V_1f+V_2f) to develop an equation relating the velocities after the collision.

Third ... use conservation of energy during an elastic collision to develop another equation relating the velocities after collision (.5m₁.713²=.5m₁V_1f²+.5m₂V_2f²).

Fourth ... solve the system of two equations to find V_1f=-.333m/s and V_2f=.310 m/s (answer a)

Fifth ... use kinematic equation v_1f²/2g = h to find the height that m₁ rebounds = .00565 m (answer b)

Sixth ... use kinematic equations .5gt² to find the amount of time it takes to fall 2 m = .639 seconds

Seventh ... use kinematic equation (d=vt) to find the distance traveled for M₂= .198m (answer c)

eighth ... realize that M1 will go back up the ramp, reverse and come back down and leave the ramp at the same velocity it had after the collision (.333 m/s).

ninth ... use kinematic equation (d=vt) to find the distance traveled for m₁ = .213 m (answer d)