Question

Two bodies of mass 1 kg and 2 kg are moving towards each other with velocity 1 ${ms}^{-1}$ and 2 ${ms}^{-1}$ respectively. If the two stick to each other after collision, the loss of energy is

• A. 6 J
• B. $4\frac{1}{2}J$
• C. 3J
• D. 2J

Answer 1

The correct option is C 3J

Conservation of momentum

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v$

$1\left(1\right)-2\left(2\right)=-\left(3\right)v$

$-3=-3v$

$v=1m/s$

$\Rightarrow$ Loss in energy $=\Delta k_2-\Delta k_1$

$=\frac{1}{2}(m_1+m_2)v^2-\frac{1}{2}{m}_1{v}_1^2-\frac{1}{2}{m}_2{v}_2^2$

$=\frac{1}{2}\times 3\times 1-\frac{1}{2}\times 1\times 1-\frac{1}{2}\times 2\times 4$

$=\frac{3}{2}-\frac{1}{2}-4$

$\Rightarrow 1.5-0.5-4$

$\Rightarrow 1-4$

$\Delta E=-3J$

$\left|\Delta E\right|=3J$

Hence $\left(c\right)$ option is correct answer