A block of mass m moving with speed v on a frictionless horizontal surface collides perfectly inelastically with an identical block (mass m ) initially at rest. The combined mass then moves onto a rou

Question

A block of mass m moving with speed v on a frictionless horizontal surface collides perfectly inelastically with an identical block (mass m ) initially at rest. The combined mass then moves onto a rough horizontal surface with coefficient of kinetic friction μ . What is the distance covered by the combined block before coming to rest?

TestHub · Accepted Answer

During the perfectly inelastic collision, momentum is conserved. Let $V$ be the velocity of the combined mass immediately after collision. $m v + m (0) = (m + m) V ⟹ m v = 2 mV ⟹ V = \frac{v}{2}$ .

After the collision, the combined mass $2 m$ moves with velocity $V$ on a rough surface. Using the work-energy theorem, the kinetic energy is dissipated by friction. $Δ K = W_{f r i c t i o n}$ $0 - \frac{1}{2} (2 m) V^{2} = - μ (2 m) g d$ Substituting $V = \frac{v}{2}$ : $- m (\frac{v}{2})^{2} = - 2 μ m g d$ $- \frac{m v ^{2}}{4} = - 2 μ m g d$ $d = \frac{m v ^{2}}{8 μ m g} = \frac{v ^{2}}{8 μg}$ .

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JEE Main 2028 (Class 11)