In the figure below, puck 1 of mass m 1 = 0.25 kg is sent sliding across a frict
ID: 2244511 • Letter: I
Question
In the figure below, puck 1 of mass m1 = 0.25 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distance2.2d from the base of the bench. What is the mass of puck 2?
kg
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 = -2.2d = V1 * t
For puck 2,
s = d = V2 * t
Therefore V1 = -2.2V2
For an elastic head-on collision, we know that the
relative velocity of approach = relative velocity of separation, so
U1 - U2= V2 - V1 = V2 - -2.2V2 = 3.2V2
Where U1 and U2 are the initial velocities of pucks 1 and 2, respectively.
But U2 = 0, so U1 = 3.2V2
Pre-collision, momentum = M1 * U1 = 0.25kg * 3.2V2
Post-collision, momentum = M1 * V1 + M2 * V2 = 0.25kg * (-2.2V2) + M2 * V2
These two are equal! Divide all terms by V2 and you have
0.25kg * 3.2 = 0.25kg * (-2.2) + M2
M2 = 0.25kg * 5.4 = 1.35 kg ? answer
Check: post-collision momentum = 0.25kg * (-2.2V2) + 1.35kg * V2 = 0.8kg * V2
pre-collision momentum = 0.25kg * 3.2V2 = 0.8kg * V2