A mass of 0.50 kg is attached to a massless spring with a spring constant k = 60

Question

A mass of 0.50 kg is attached to a massless spring with a spring constant k = 600 N/m (see figure). The system rests on a level, friction-free surface and is initially at rest. A second mass of 0.20 kg makes an elastic head-on collision with the mass attached to the spring; thereafter, the oscillating system vibrates with an amplitude of 0.25 m.

A.What was the incident speed of the second mass (Ans - 16m/s needhelp with work)

B.What is the maximum value of the potential energy stored in the spring during the subsequent oscillations?

Explanation / Answer

A ) conserving energy, all the K.E of the mass m2 will be converted to spring potential energy and some part will remain as K.E of m2, so, .5*m2*v^2 = .5*k*A^2 + .5*m2*v'^2 where v' = final velocity of m2 after collision so, using momentum conservation, m2*v = m1*u + m2*v' and all the K.E imparted to the m1 after collision is converted to spring P.E so, .5*m1*u^2 = .5*k*A^2 so, u = 8.67 m/s so, v' = (0.2*v - 0.5*8.67)/0.2 substituting v' in the 1st equation and solving the quadratic equation for v, we get, v = 15.16 m/s

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