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 5th 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 ________.
Step 1: Understanding the Concept:
The distance covered in the $n^{th}$ second is given by the formula for uniformly accelerated motion. We first find the acceleration for both masses, then calculate their final velocities at $t = 10$ s to find the ratio of their momenta ($p = mv$). Step 2: Key Formula or Approach:
1. Distance in $n^{th}$ second: $S_n = u + \frac{a}{2}(2n - 1)$.
2. Velocity at time $t$: $v = u + at$.
3. Momentum: $p = mv$. Step 3: Detailed Explanation:
For Mass 1 ($m_1 = 3.4$ kg, $u_1 = 5$ m/s, $S_5 = 104$ m):
\[ 104 = 5 + \frac{a_1}{2}(2(5) - 1) \implies 99 = \frac{9a_1}{2} \implies a_1 = 22 \text{ m/s}^2 \]
Velocity after 10s: \( v_1 = 5 + 22(10) = 225 \text{ m/s} \).
Momentum \( p_1 = 3.4 \times 225 = 765 \text{ kg}\cdot\text{m/s} \).
For Mass 2 ($m_2 = 2.5$ kg, $u_2 = 12$ m/s, $S_5 = 129$ m):
\[ 129 = 12 + \frac{a_2}{2}(9) \implies 117 = \frac{9a_2}{2} \implies a_2 = 26 \text{ m/s}^2 \]
Velocity after 10s: \( v_2 = 12 + 26(10) = 272 \text{ m/s} \).
Momentum \( p_2 = 2.5 \times 272 = 680 \text{ kg}\cdot\text{m/s} \).
Ratio of momenta:
\[ \frac{p_1}{p_2} = \frac{765}{680} = \frac{153 \times 5}{136 \times 5} = \frac{153}{136} \]
Dividing by 17:
\[ \frac{153 \div 17}{136 \div 17} = \frac{9}{8} \]
Comparing with $\frac{x}{8}$, we find $x = 9$. (Note: Calculation verification for $x=17$ depends on specific mass/velocity pairings; for these values, $x=9$). Step 4: Final Answer:
The value of $x$ is 9.
