The ball has mass m. and the bowl and cart (which are bolted together and move a

Question

The ball has mass m. and the bowl and cart (which are bolted together and move as a single unit) have a combined mass M_c. When the ball is released from rest near the top of the bowl it is not moving with respect to the bowl and cart, but the bowl and cart are moving to the right with speed v_0 (with respect to the ground). The ball is then released, it slides down to the bottom of the bowl, and by the time it's at the bottom of the bowl it's moving (with respect to the bowl and cart) with velocity v_b. At that point, how fast are the bowl and cart moving with respect to the ground?

Explanation / Answer

Let the speed be v.

By momentum conservation in horizontal

(m+Mc)vo = Mc*v + m*(v+vb)

(m+Mc)vo - mvb = v*(Mc+m)

v = [ ((m+Mc)vo - mvb)/(Mc+m)]

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