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Two blocks of mass $2 m$ and $m$ slide down the slope $A B$ having same friction coefficient. Then :

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The heavier mass reaches the bottom in a shorter time

The heavier mass has twice the momentum than that of the other when reaching B

The lighter mass reaches the bottom in a short time

Both the masses have the same kinetic energy on arriving at B

answer is C.

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Detailed Solution

$a = g \sin θ - μ g \cos θ$ is same for $(1)$ and $(2)$ both will read bottom with same speed $u^{2} = 0 + 2 as$

$k_{1} = \frac{1}{2} (2 m) v_{1}^{2} k_{2} = \frac{1}{2} (m) v_{2}^{2} k_{1} \neq k_{2} p_{1} = 2 {mv}_{1}; p_{2} = {mv}_{2} p_{1} = 2 p_{2}$

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