A steel ball of mass 0.400 kg is fastened to a cord that is 60.0 cm long and fix
ID: 1325940 • Letter: A
Question
A steel ball of mass 0.400 kg is fastened to a cord that is 60.0 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 3.00 kg steel block initially at rest on a frictionless surface. The collision is elastic.
(a) Find the speed of the ball just after collision.
m/s
(b) Find the speed of the block just after collision.
m/s
Explanation / Answer
? the total energy at the collision is pot energy
E =m1*gh, where m1=0.4 kg, h=0.6 m;
? thus kin energy of the ball just before collision
E=0.5m1*v^2; thence v=?(2gh);
? now after collision: momentum conserv law says
m1*v1+m2*v2 =m1*v, where m2=3kg, v1 is speed of the ball, v2 is speed of the block; thence v2=m1*(v-v1)/m2;
and energy conserv law says
0.5m1*v1^2 +0.5m2*v2^2 = E; or;
m1*v1^2 +m2*{m1*(v-v1)/m2}^2 = 2m1*gh; or;
v1^2 +(m1/m2)*{v^2 -2v*v1 +v1^2} =2gh; or;
? (m1+m2)*v1^2
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