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Two blocks of mass m and 2m connected by a weightless spring of stiffness k rest on a smooth horizontal plane. Block of mass m is shifted to a small distance x to the left and then released. If the duration between releasing mass m and breaking off of mass 2m is $∆$ t , then average force(in newton) exerted by wall on mass 2m is given as $\frac{x}{Δt} \sqrt{mkn}$ . Find the value of n.

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Detailed Solution

As impulse exerted by wall is equal to change in momentum of the system
$F_{av} \cdot Δt = 3 m \cdot V_{cm} = 3 m \cdot \frac{x}{3} \sqrt{\frac{k}{m}} \Rightarrow F_{av} = \frac{x}{Δt} \sqrt{km} = \frac{x}{Δt} \sqrt{km \cdot n} (given) ∴ n = 1$

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