Momentum and CollisionHard
Question
Figure shows a block A of mass 5 kg kept at rest on a horizontal smooth surface. A spring (K = 200 N/m) which is compressed by 10 cm and tied with the help of a string to maintain the compression is attached to block A as shown in figure. Block B also of mass 5 kg moving with 2 m/s collides with A, as shown. During the collision the string breaks and after the collision the spring is in its natural state. Assume the bodies to be elastic and let the velocities of A and B be v1 and v2 respectively assuming positive direction towards right, after collision. Then
Options
A.v1 + v2 > 2
B.Initial kinetic energy of system = final kinetic energy of system
C.v12 + v22 = 4.4 (m/s)2
D.v1 - v2 = 2
Solution
Energy conservation 
mv12 +
mv22 = 
mu2 + 
kx2 ⇒ v12 + v22 = 4.4(m/s)2
mv12 + mv22 = mu2 + kx2 ⇒ v12 + v22 = 4.4(m/s)2Create a free account to view solution
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