Two masses of 3.4 kg and 2.5 kg are…

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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 5th 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 in the $n$ th second: $s_n = u + \frac{a}{2}(2n - 1)$ . For $n = 5$ , $s_5 = u + 4.5a$ .

First mass: $104 = 5 + 4.5a_1 \Rightarrow a_1 = 22$ m/s².
Second mass: $129 = 12 + 4.5a_2 \Rightarrow a_2 = 26$ m/s².

After 10 s: $v_1 = 5 + 220 = 225$ m/s and $v_2 = 12 + 260 = 272$ m/s.

$\frac{p_1}{p_2} = \frac{3.4 \times 225}{2.5 \times 272} = \frac{765}{680} = \frac{9}{8}.$

Answer: $x = 9$ .

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Kinematics

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