Bob sees a stationary cart 8.0 m in front of him. He decides to run to the cart

Question

Bob sees a stationary cart 8.0 m in front of him. He decides to run to the cart as fast as he can, jump on, and roll down the street. Bob has a mass of 75 kg and the cart's mass is 25 kg. If Bob accelerates at a steady 1.0 m/s^2, what is the cart's speed just after Bob jumps on? A 238 U uranium nucleus is radioactive. It spontaneously disintegrates into a small fragment that is ejected with a measured speed of 1.50 * 107 m/s and a "daughter nucleus" that recoils with a measured speed of 2.56 * 105 m/s. What are the atomic masses of the ejected fragment and the daughter nucleus?

Explanation / Answer

a = 1 m/s^2 ; S = 8 m ; m1 = 75 kg ; m2 = 25 kg

Let vf be the required speed.

We know from eqn of motion

v^2 = u^2 + 2 a S

since u = 0 in our case, v1 = sqrt (2 a S)

v1 = sqrt (2 x 1 x 8) = 4 m/s

We know from conservation of momentum

Pi = Pf

m1 v1 + m2 v2 = (m1 + m2) vf

v2 = 0 since the cart is stationary

vf = m1v1/(m1 + m2)

vf = 75 x 4/(75 + 25) = 3 m/s

Hence, vf = 3 m/s

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