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 $$\frac{x}{8}$$. The value of $$x$$ is __________.

Solution :

For uniformly accelerated motion, distance travelled in \(n^{th}\) second is :

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

For first mass :

$$u_1 = 5\text{ m s}^{-1}$$

Distance in 5th second :

$$104 = 5 + \frac{a_1}{2}(9)$$

$$99 = \frac{9a_1}{2}$$

$$a_1 = 22\text{ m s}^{-2}$$

Velocity after 10 s :

$$v_1 = u_1 + a_1t$$

$$= 5 + 22(10)$$

$$= 225\text{ m s}^{-1}$$

Momentum :

$$p_1 = m_1v_1$$

$$= 3.4 \times 225$$

$$= 765$$

For second mass :

$$u_2 = 12\text{ m s}^{-1}$$

Distance in 5th second :

$$129 = 12 + \frac{a_2}{2}(9)$$

$$117 = \frac{9a_2}{2}$$

$$a_2 = 26\text{ m s}^{-2}$$

Velocity after 10 s :

$$v_2 = u_2 + a_2t$$

$$= 12 + 26(10)$$

$$= 272\text{ m s}^{-1}$$

Momentum :

$$p_2 = m_2v_2$$

$$= 2.5 \times 272$$

$$= 680$$

Therefore,

$$\frac{p_1}{p_2} = \frac{765}{680}$$

$$= \frac{9}{8}$$

Given ratio is :

$$\frac{x}{8}$$

Hence,

$$x = 9$$

Final Answer :

$$9$$