When a moving body collides with a stationary body of \(n\) times its mass, then the amount of kinetic…
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
\(\frac{4 n}{(1+n)^2}\)
\(\frac{n}{(1+n)^2}\)
\(\frac{n^2}{(1+n)^2}\)
\(\frac{4 n^2}{(1+n)^2}\)
Solution
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,
\(\begin{aligned}
m_1 u & =m_1 v_1+M_2 v_2 \\
\Rightarrow m_1 u & =m_1 v_1+n m_1 v_2 \\
u & =v_1+n v_2 \quad \ldots (ii) \\
u & =v_2-u+n v_2 \quad \text{[From Eq. (i)]} \\
\Rightarrow 2 u & =(n+1) v_2 \\
v_2 & =\frac{2 u}{n+1}
\end{aligned}\)
Putting this value in Eq. (i)
\(v_1=v_2-u=\frac{2 u}{n+1}-u=\frac{2 u-n u-u}{n+1}=\frac{(1-n) u}{n+1}\)
\(\therefore\) Kinetic energy of moving body of mass \(m_1\), before collision,
\(K_1=\frac{1}{2} m_1 u^2\)
Kinetic energy of stationary body after collision,
\(\begin{aligned}
K_2 & =\frac{1}{2} M_2 v_2^2=\frac{1}{2} n m_1\left(\frac{2 u}{n+1}\right)^2 \\
& =\frac{1}{2} m_1 u^2 \cdot \frac{4 n}{(n+1)^2}
\end{aligned}\)
\(\therefore\) Amount of KE transferred to stationary body
\(=\frac{K_2}{K_1}=\frac{\frac{1}{2} m_1 u^2 \times \frac{4 n}{(n+1)^2}}{\frac{1}{2} m_1 u^2}=\frac{4 n}{(n+1)^2}\)