Two particles of mass m1 and m2 are moving with velocity \left(\overset{\rightarrow}{v}\right)_{1} and \left(\overset{\rightarrow}{v}\right)_{2} respectively. Mark the correct statement about a perfectly inelastic collision between m1 and m2.

Engineering

Physics

Center of Mass

Collision

Momentum and Energy Conservation for System of Particles

Question

Two particles of mass m₁ and m₂ are moving with velocity ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively. Mark the correct statement about a perfectly inelastic collision between m₁ and m₂.

Impulse by m₁ on m₂ in CM frame is $\frac{m_{1} m_{2}}{m_{1} + m_{2}} ({\vec{v}}_{1} - {\vec{v}}_{2})$ .

Momentum transfered by m₂ to m₁ is $\frac{m_{1} m_{2}}{m_{1} + m_{2}} ({\vec{v}}_{2} - {\vec{v}}_{1})$ .

In CM frame both the particles come to rest after perfectly inelastic collision.

Impulse by m₁ to m₂ is $\frac{m_{1} m_{2}}{m_{1} + m_{2}} ({\vec{v}}_{1} - {\vec{v}}_{2})$ .

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In a perfectly inelastic collision, the two particles stick together after collision. The impulse on a body equals its change in momentum. In the center of mass (CM) frame, the total momentum is zero. The impulse by m₁ on m₂ is equal to the change in momentum of m₂. The reduced mass, μ, is given by $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}$ . The relative velocity is ${\vec{v}}_{1} - {\vec{v}}_{2}$ . The impulse is $μ ({\vec{v}}_{1} - {\vec{v}}_{2})$ .

Final Answer: The first and fourth options are correct.