JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Kinematics

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 ______.

Answer: 9

Solution

The distance covered in the

n

th second is

S_{n} = u + \frac{a}{2} (2 n - 1)

. For the first body,

104 = 5 + \frac{a _{1}}{2} (9)

gives

a_{1} = 22 m/s^{2}

; for the second,

129 = 12 + \frac{a _{2}}{2} (9)

gives

a_{2} = 26 m/s^{2}

. After

10 s

the speeds are

v_{1} = 5 + 22 \times 10 = 225 m/s

and

v_{2} = 12 + 26 \times 10 = 272 m/s

. The momenta are

p_{1} = 3.4 \times 225 = 765

and

p_{2} = 2.5 \times 272 = 680

, so

\frac{p _{1}}{p _{2}} = \frac{765}{680} = \frac{9}{8}

. The value of

x

9

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer

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