A steel ball of mass 0.680 kg is fastened to a cord that is 34.0cm long and fixe

Question

A steel ball of mass 0.680 kg is fastened to a cord that is 34.0cm long and fixed at the far end. The ball is then released whenthe cord is horizontal (Fig. 9-68). At the bottom of its path, theball strikes a 4.00 kg steel block initially at rest on africtionless surface. The collision is elastic. Find(a) the speed of the ball and (b)the speed of the block, both just after the collision.

Explanation / Answer

initial PE of the ball = mgh just before collision, all PE is converted to KE. so if the speedof the ball is u just before collision, (1/2)mu2 = mgh u = (2gh) >>> compute this first. after the collision, let the ball speed be v and the block speed beV. the mass of the block is M. since the collision is elastic, both momentum and KE are conserved.so mu = mv + MV m(u-v) = MV >>> equation A (1/2)mu2 = (1/2)mv2 + (1/2)MV2 m(u2 - v2) = MV2 >>>equation B dividing equation B by A, u+v = V >>> equation C also from equation A, u-v = (M/m)V >>> equation D so we get, from C and D, u = (1+M/m)V/2 V = 2u/(1+M/m) v = V(1-M/m) = 2u(1-M/m)/(1+M/m) since you know u, you can compute both v and V. hope this helps!

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