Q75 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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 $5^\text{th}$ second are $104$ m and $129$ m, respectively. The ratio of their momenta after $10$ s is $\dfrac{x}{8}$. The value of $x$ is ________.
Correct answer: 9
Solution
The distance traversed in the $n^{\text{th}}$ second is given by $S_n = u + \dfrac{a}{2}(2n - 1)$.
For the first mass ($m_1 = 3.4$ kg, $u_1 = 5$ m/s):
$104 = 5 + \dfrac{a_1}{2}(2 \times 5 - 1)$
$99 = \dfrac{9a_1}{2} \Rightarrow a_1 = 22$ m/s$^2$
For the second mass ($m_2 = 2.5$ kg, $u_2 = 12$ m/s):
$129 = 12 + \dfrac{a_2}{2}(2 \times 5 - 1)$
$117 = \dfrac{9a_2}{2} \Rightarrow a_2 = 26$ m/s$^2$
Velocities of the masses after $10$ s:
$v_1 = u_1 + a_1 t = 5 + 22 \times 10 = 225$ m/s
$v_2 = u_2 + a_2 t = 12 + 26 \times 10 = 272$ m/s
Momenta of the masses after $10$ s:
$p_1 = m_1 v_1 = 3.4 \times 225 = 765$ kg m/s
$p_2 = m_2 v_2 = 2.5 \times 272 = 680$ kg m/s
Ratio of their momenta:
$\dfrac{p_1}{p_2} = \dfrac{765}{680} = \dfrac{9}{8}$
Comparing with $\dfrac{x}{8}$, we get $x = 9$.
Answer: $9$
For the first mass ($m_1 = 3.4$ kg, $u_1 = 5$ m/s):
$104 = 5 + \dfrac{a_1}{2}(2 \times 5 - 1)$
$99 = \dfrac{9a_1}{2} \Rightarrow a_1 = 22$ m/s$^2$
For the second mass ($m_2 = 2.5$ kg, $u_2 = 12$ m/s):
$129 = 12 + \dfrac{a_2}{2}(2 \times 5 - 1)$
$117 = \dfrac{9a_2}{2} \Rightarrow a_2 = 26$ m/s$^2$
Velocities of the masses after $10$ s:
$v_1 = u_1 + a_1 t = 5 + 22 \times 10 = 225$ m/s
$v_2 = u_2 + a_2 t = 12 + 26 \times 10 = 272$ m/s
Momenta of the masses after $10$ s:
$p_1 = m_1 v_1 = 3.4 \times 225 = 765$ kg m/s
$p_2 = m_2 v_2 = 2.5 \times 272 = 680$ kg m/s
Ratio of their momenta:
$\dfrac{p_1}{p_2} = \dfrac{765}{680} = \dfrac{9}{8}$
Comparing with $\dfrac{x}{8}$, we get $x = 9$.
Answer: $9$
Hence, the correct answer is 9.