Question

A block of mass $0.50kg$ is moving with a speed of $2.00m{s}^{-1}$ on a smooth surface. It strikes another mass of $1.00kg$ which is at rest and then they move together as a single body. The energy which is lost during the collision is

• A. 0.16 J
• B. 1.00 J
• C. 0.67 J
• D. 0.34 J

Answer 1

Answer: (C)

Here, $m_1=0.50kg,u_1=2.00m{s}^{-1}$
$m_2=1.00kg,u_2=0$
If $v$ is the velocity of the combination after collision, then according to the principle of conservation of linear momentum,
$m_1{u}_1+m_2{u}_2=\left(m_1+m_2\right)v$
or $v=\frac{m_1{u}_1}{m_1+m_2}=\frac{0.50\times 2.00}{0.50+1.00}=\frac{1.00}{1.50}=\frac{2}{3}m{s}^{-1}\left(\because u_2=0\right)$
Energy loss = Initial energy - Final energy
$=\frac{1}{2}{m}_1{u}_1^2-\frac{1}{2}\left(m_1+m_2\right)v^2$
$=\frac{1}{2}\times 0.50\times (2.00{)}^2-\frac{1}{2}\times (0.50+1.00)\times {\left(\frac{2}{3}\right)}^2$
$=1.00-\frac{1.50}{2}\times \frac{4}{9}=0.67J$