, 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

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

that tool is called an altimeter which measures your altitude which can be used to tell which atmosphere you are currently in, including the stratosphere.

hope this ! : )