Puck 1 of mass m1 is sent sliding across a frictionless lab bench, to undergo a

Question

Puck 1 of mass m1 is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 is then ejected horizontally off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and is ejected horizontally off the opposite edge of the bench, landing a distance 6d from the base of the bench. What is the mass of puck 2, m2, in terms of m1, d, and g, or a subset of these quantities?

Puck 1 of mass m, is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 is then ejected horizontally -- s tha rieaned haoisamnal off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and is ejected horizontally off the opposite edge of the bench, landing a distance 6d from the base of the bench. What is the mass of puck 2, m2, in terms of mi, d, and g, or a subset of these quantities?

Explanation / Answer

When the pucks slid off the table, they both fell for the same time "t".
Puck 1 had horizontal velocity V1, and the horizontal distance it traveled was
s = -6d = V1 * t
For puck 2,
s = d = V2 * t
Therefore V1 = -6V2

For an elastic head-on collision, we know that the
relative velocity of approach = relative velocity of separation, so
U1 - U2 = V2 - V1 = V2 - (-6V2) = 7V2
Where U1 and U2 are the initial velocities of pucks 1 and 2, respectively.
But U2 = 0, so U1 = 7V2

Pre-collision, momentum = M1 * U1 = M1 * 7V2
Post-collision, momentum = M1 * V1 + M2 * V2 = M1* (-6V2) + M2 * V2
These two are equal! Divide all terms by V2 and you have
M1* 7 = M1 * (-6) + M2
M2 = 13M1

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