In a smooth stationary cart of length d,a sma | vedclass

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(D) Let $m$ be the mass of the block and $M$ be the mass of the cart. Since the system is on a smooth surface,the momentum of the system is conserved.

Vedclass · Accepted Answer

infinite

Vedclass · Answer

(D) Let $m$ be the mass of the block and $M$ be the mass of the cart. Since the system is on a smooth surface,the momentum of the system is conserved.Initially,the block has velocity $v$ and the cart is at rest. After the first collision,the block and the cart move with some velocities $v_1$ and $u_1$ respectively. The relative velocity of separation is $e$ times the relative velocity of approach.In each collision,the relative velocity of the block with respect to the cart decreases by a factor of $e$. The time taken for the first collision is $t_1 = d/v$.After the first collision,the relative velocity becomes $ev$. The time taken for the second collision is $t_2 = d/(ev)$.After the second collision,the relative velocity becomes $e^2v$. The time taken for the third collision is $t_3 = d/(e^2v)$,and so on.The total time $T$ is the sum of the times between successive collisions: $T = \frac{d}{v} + \frac{d}{ev} + \frac{d}{e^2v} + \dots$This is a geometric progression with the first term $a = d/v$ and common ratio $r = 1/e$.Since $e \leq 1$,the common ratio $r = 1/e \geq 1$.$A$ geometric series with $r \geq 1$ diverges to infinity.Therefore,the time taken for the block to come to rest with respect to the cart is infinite.

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