Question

A block of mass $m_1$ undergoes a completely inelastic collision with an initially stationary block of mass $m_2$. Let K be the kinetic energy transferred to $m_2$ after collision and $K_1$ be the initial kinetic energy. Consider all motion to be in one dimension. The ratio $K/K_1$

• A. is constant for all values of $m_2/m_1$
• B. attains maximum value when $m_2/m_1=1$
• C. decreases for $m_2/m_1>1$
• D. has a maximum value of 1/4

Answer 1

Answer: (B), (C), (D)

The correct options are
B attains maximum value when $m_2/m_1=1$
C decreases for $m_2/m_1>1$
D has a maximum value of 1/4

In completely inelastic collision, both masses $m_1$ and $m_2$ stick to each other.

Let, initial speed of $m_1=u$ and the speed of the system after collision be $v$.

Applying conservation of momentum:-

$m_{1}u=(m_1+m_2)v$

$⟹v=\frac{m_{1}u}{m_1+m_2}$

$K_1=\frac{1}{2}{m}_1{u}^2$

And, $K=\frac{1}{2}{m}_2{v}^2=\frac{1}{2}\frac{m_2{m}_1^2{u}^2}{(m_1+m_2{)}^2}$

$⟹\frac{K}{K_1}=\frac{m_2{m}_1}{(m_1+m_2{)}^2}$

$⟹\frac{K}{K_1}=\frac{\frac{m_2}{m_1}}{{(1+\frac{m_2}{m_1})}^2}$

Let, $x=\frac{m_2}{m_1}$ and $\frac{K}{K_1}=\frac{x}{(1+x{)}^2}=f\left(x\right)\left(say\right)$

Now, $f^{{}^{\prime }}\left(x\right)=\frac{(1+x{)}^2-2x(1+x)}{(1+x^2{)}^2}$

$f^{{}^{\prime }}\left(x\right)=\frac{1-x^2}{(1+x^2{)}^2}$

Now, $f^{{}^{\prime }}\left(x\right)=0$ for $x=1$ and $f^{{}^{\prime }}\left(x\right)<0\forall x>1$

Hence, $\frac{k}{K_1}=f\left(x\right)$ attains its maximum value at $x=\frac{m_2}{m_1}=1$

And, maximum value is $f\left(1\right)=\frac{1}{4}$

$\frac{k}{K_1}=f\left(x\right)$ decreases for $x=\frac{m_2}{m_1}>1$

Hence, (B),(C), and (D) are correct options.