Question

A block of mass $m_1=0.5\text{kg}$ is moving with the speed of $u_1=2\text{m/s}$ on a smooth surface. It strikes another mass of $m_2=1\text{kg}$ initially at rest and then they move together as a single body. The energy lost 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 $0.67\text{J}$

Before collision

After collision



By the law of conservation of linear momentum
$m_1{u}_1=(m_1+m_2)v$
$0.5\times 2=(1+0.5)v\Rightarrow v=\frac{1}{1.5}$
Initial K.E. of system$(m_1+m_2)$
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$

$=\frac{1}{2}\times 0.5\times (2{)}^2+\frac{1}{2}\times 1\times 0$

$=1\text{J}$

Final K.E. of system$(m_1+m_2)$
$\frac{1}{2}(m_1+m_2)v^2$
$=\frac{1}{2}\times 1.5\times {\left(\frac{1}{1.5}\right)}^2=\frac{1}{3}=0.33\text{J}$
$\therefore$ Loss of energy $=1-0.33=0.67\text{J}$