AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
When a moving body collides with a stationary body of \(n\) times its mass, then the amount of kinetic energy transferred to the stationary body is
- A \(\frac{4 n}{(1+n)^2}\)
- B \(\frac{n}{(1+n)^2}\)
- C \(\frac{n^2}{(1+n)^2}\)
- D \(\frac{4 n^2}{(1+n)^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 n}{(1+n)^2}\)
Step-by-step Solution
Detailed explanation
Suppose, mass of moving body is \(m_1\). Mass of stationary body, \(M_2=n m_1\) For elastic collision, velocity of separation = velocity of approach \(\begin{aligned} v_2-v_1 & =u-0 \\ v_2-u & =v_1 \quad \ldots (i) \end{aligned}\) By the law of conservation of momentum,…
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