Calculate Parameter x from Momentum Ratio of Accelerated Masses

To find the accelerations $a_1$ and $a_2$ of the two masses, we use the formula for distance covered in the $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$

For the first mass ($m_1 = 3.4\text{ kg}$, $u_1 = 5\text{ m/s}$): $104 = 5 + \frac{a_1}{2}(2 \cdot 5 - 1)$ $99 = \frac{9}{2} a_1 \implies a_1 = 22\text{ m/s}^2$

For the second mass ($m_2 = 2.5\text{ kg}$, $u_2 = 12\text{ m/s}$): $129 = 12 + \frac{a_2}{2}(2 \cdot 5 - 1)$ $117 = \frac{9}{2} a_2 \implies a_2 = 26\text{ m/s}^2$

Next, we calculate their velocities after $t = 10\text{ s}$ using $v = u + at$: $v_1 = 5 + 22(10) = 225\text{ m/s}$ $v_2 = 12 + 26(10) = 272\text{ m/s}$

The momenta $p_1$ and $p_2$ after $10\text{ s}$ are: $p_1 = m_1 v_1 = 3.4 \times 225 = 765\text{ kg}\cdot\text{m/s}$ $p_2 = m_2 v_2 = 2.5 \times 272 = 680\text{ kg}\cdot\text{m/s}$

Taking the ratio of their momenta: $\frac{p_1}{p_2} = \frac{765}{680} = \frac{9}{8}$

Given that $\frac{p_1}{p_2} = \frac{x}{8}$, we get: $x = 9$