Two masses of $3.4$ kg and $2.5$ kg are accelerated from an initial speed of $5\ \text{m/s}$ and $12\ \text{m/s}$, 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 ______.
Numerical value type. Enter your answer.
Answer: 9
Distance in the $n$th second: $s_n = u + \dfrac{a}{2}(2n - 1)$. For $n = 5$: $s_5 = u + 4.5a$.
Mass 1: $104 = 5 + 4.5a_1 \Rightarrow a_1 = 22$ m/s². Mass 2: $129 = 12 + 4.5a_2 \Rightarrow a_2 = 26$ m/s².
After $10$ s: $v_1 = 5 + 220 = 225$ m/s, $v_2 = 12 + 260 = 272$ m/s.
$$\frac{p_1}{p_2} = \frac{3.4 \times 225}{2.5 \times 272} = \frac{765}{680} = \frac{9}{8}$$
$x = 9$.
Solution by Sreeraj P, M.Sc Physics