JEE Mains · Physics · STD 11 - 2. motion in straight line
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 ________.
- A 10
- B 8
- C 9
- D 12
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
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…
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