A steel ball of mass 0.400 kg is fastened to a cord that is 70.0 cm long and fix

Question

A steel ball of mass 0.400 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic.

(a) Find the speed of the ball just after collision.
1 m/s

(b) Find the speed of the block just after collision.
2 m/s

Explanation / Answer

YOUR GIVEN ANSWERE IS REVERSED WHEN YOU TYPING... ****************************************************************************************************** the mass of the ball = m = 0.400 kg mass of the block = M = 2.5 kg length of the cord = 70 cm = .70 m initial speed of the block = u2 = 0 initial speed of the ball = u1= u = v(2gh) = 3.70 m/s ........... A] in an elastic collision, both momentum and KE are conserved.so (1/2)mv2 = (1/2)mu2 +(1/2)MV2 >>> u = final velocity of the ballafter collision, M=mass of block, V=velocity of block after collision mv = mu + MV inserting V from this equation into energy equation, mv2 = mu2 + M(m(v-u)/M)2 v2 = u2 + m(v-u)2/M v+u = m(v-u)/M v= (m-M) * u/(m+M) = [ .4-2.5]* 3.7 / [ .4 + 2.5] = 2 m/s B] Find the speed of the block just after collision. = 2mu/[m+M] = 1 m /s

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