In the figure below, puck 1 of mass m 1 = 0.15 kg is sent sliding across a frict
ID: 2256659 • Letter: I
Question
In the figure below, puck 1 of mass m1 = 0.15 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 distance 2.3d from the base of the bench. What is the mass of puck 2?
? kg
Explanation / Answer
Let the velocities of 1 and 2 be : V1 and V2 after the collision
velocity of 1 before collision = v
m1*v = m2*V2 - m1*V1
from elastic collision: v = V1 + V2
since landing distance is directly proportional to the velocity.
therefore: V1 = 2.3V2
m1(v + 2.3V2) = m2 V2
simplify to get:
m2 = 0.84 kg