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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

  1. A \(\frac{4 n}{(1+n)^2}\)
  2. B \(\frac{n}{(1+n)^2}\)
  3. C \(\frac{n^2}{(1+n)^2}\)
  4. D \(\frac{4 n^2}{(1+n)^2}\)
Verified Solution

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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