Question

A body of mass 2kg is placed on a horizontal frictionless surface. It is connected to one end of a spring whose force constant is $250N/m$. The other end of the spring is joined with the wall. A particle of mass 0.15kg moving horizontally with speed $v$ sticks to the body after collision. If it compresses the spring by $10cm$, the velocity of the particle is

• A. $3m/s$
• B. $5m/s$
• C. $10m/s$
• D. $15m/s$

Answer 1

Answer: (D)

$15m/s$

Answer 2

By the conservation of momentum

Initial momentum of particle = Final momentum of system ⇒ m × v = (m + M) V

∴ velocity of system $V = \frac{mv}{(m + M)}$

Now the spring compresses due to kinetic energy of the system so by the conservation of energy

$\frac{1}{2}kx^{2} = \frac{1}{2}(m + M)V^{2} = \frac{1}{2}(m + M)\left( \frac{mv}{m + M} \right)^{2}$

⇒ $kx^{2} = \frac{m^{2}v^{2}}{m + M} \Rightarrow v = \sqrt{\frac{kx^{2}(m + M)}{m^{2}}} = \frac{x}{m}\sqrt{k(m + M)}$

Putting m = 0.15 kg, M = 2 kg, k = 250 N/m, x = 0.1 m we get v = 15 m/s.