JEE Main 2026 Apr 8 Shift 2, Physics Q25: Motion In One Dimension

JEE Main2026Apr 8, Shift 2Physics

Two masses of $3.4$ kg and $2.5$ kg are accelerated from an initial speed of $5$ m/s and $12$ m/s, respectively. The distances traversed by the masses in the $5^{th}$ second are $104$ m and $129$ m, respectively. The ratio of their momenta after $10$ s is $\frac{x}{8}$ . The value of $x$ is _________.

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Solution

Distance covered in the $n^{th}$ second: $S_{n} = u + \frac{a}{2} (2 n - 1)$ .

For $n = 5$ , $2 n - 1 = 9$ .

Mass 1: $104 = 5 + \frac{a _{1}}{2} (9) ⟹ a_{1} = \frac{2 \cdot 99}{9} = 22$ m/s $^{2}$ .

Mass 2: $129 = 12 + \frac{a _{2}}{2} (9) ⟹ a_{2} = \frac{2 \cdot 117}{9} = 26$ m/s $^{2}$ .

Velocity after $10$ s:

$v_{1} = 5 + 22 (10) = 225$ m/s, $v_{2} = 12 + 26 (10) = 272$ m/s.

Momentum ratio:

$\frac{p _{1}}{p _{2}} = \frac{( 3.4 ) ( 225 )}{( 2.5 ) ( 272 )} = \frac{765}{680} = \frac{9}{8}$ .

$∴$ $x = 9$ .

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JEE Main 2026 Apr 8 Shift 2, Physics Q25: Motion In One Dimension

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