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 x/8 . The value of x is

Physics · Motion in one Dimension

JEE Main 2026 — 8 April, Evening Shift — Question 20

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

Answer: 9

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Step-by-step solution

Using $s_n = u + \frac{a}{2}(2n-1)$ : For first mass: $104 = 5 + \frac{a_1}{2}\times9$ ⇒ $a_1 = 22$ m/s². For second: $129 = 12 + \frac{a_2}{2}\times9$ ⇒ $a_2 = 26$ m/s². Velocity after 10 s: $v_1 = 5+22\times10=225$ m/s, $v_2 = 12+26\times10=272$ m/s. Momentum ratio: $\frac{p_1}{p_2} = \frac{3.4\times225}{2.5\times272} = \frac{765}{680} = \frac{9}{8}$ . Thus $x=9$ .

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion

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Answer: 9

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