, puck 1 of mass m1 ! 0.20 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 2d from the base of the bench. What is the mass of puck 2

1 kg

Explanation:

Assuming that,

Δx(2) = v(2)t, where Δx(2) = d and v(2) = 2m1 / (m1 + m2) v1i

On the other hand again, if we assume that

Δx(1) = v(1)t, where Δx(1) = -2d, and v(1)t = m1 - m2 / m1 + m2 v1i

From the above, we proceed to dividing Δx(2) by Δx(1), so that we have

d/-2d = [2m1 / (m1 + m2) v1i] / [m1 - m2 / m1 + m2 v1i], this is further simplified to

1/-2 = [2m1 / (m1 + m2)] / [m1 - m2 / m1 + m2]

1/-2 = 2m1 / (m1 + m2) * m1 + m2 / m1 - m2

1/-2 = 2m1 / m1 - m2, if we cross multiply, we have

m1 - m2 = -2 * 2m1

m1 - m2 = -4m1

m2 = 5m1

From the question, we're told that m1 = 0.2 kg, if we substitute for that, we have

m2 = 5 * 0.2

m2 = 1 kg

2b2t

explanation:

2b2t

the answer is b, have negligible mass