Question

A block of mass $0.50\text{kg}$ is moving with a speed of $2.00{\text{ms}}^{-1}$ on a smooth surface. it strikes another mass of $1.00\text{kg}$ and then they move together as a single body. The energy loss during the collision is

• A. $0.16\text{J}$
• B. $1.00\text{J}$
• C. $0.67\text{J}$
• D. $0.34\text{J}$

Answer 1

The correct option is C $0.67\text{J}$

From law of conservation of momentum, we have
$m_1{v}_1+m_2{v}_2=(m_1+m_2)v$
Given, $m_1=0.50\text{kg},v_1=2{\text{ms}}^{-1}$,
$m_2=1\text{kg},v_2=0\left[\text{at rest}\right]$
$0.5\times 2+1\times 0=1.5\times v$
[assumed that 2nd body is at rest]
$\Rightarrow v=\frac{2}{3}$
$\therefore \Delta K=K_f-K_i$
$\frac{1.5\times {\left(\frac{2}{3}\right)}^2}{2}-\left(0.5\right)\times \frac{2^2}{2}=-\frac{2}{3}J$
$=-0.67\text{J}$
So, energy lost is $0.67\text{J}$.