Question

A block of mass $2\text{kg}$ moving at $2{\text{ms}}^{-1}$ collides head on inelastically with another block of equal mass kept at rest. Find the maximum loss in kinetic energy due to the collision.

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

Answer 1

The correct option is B $2\text{J}$

Given, $m_1=m_2=m=2\text{kg};v_1=2{\text{ms}}^{-1};v_2=0{\text{ms}}^{-1}$





Since, the net external force on the system is zero, so using conservation of linear momentum, we have

$P_i=P_f$

$\Rightarrow m\times 2+m\times 0=(m+m)v$

$\Rightarrow v=1{\text{ms}}^{-1}$

Maximum loss in kinetic energy,

$K.E_{loss}=K.E_i-K.E_f....\left(1\right)$

The initial kinetic energy of the system is,

$K.E_i=\frac{1}{2}\times 2\times (2{)}^2+\frac{1}{2}\times (2)\times (0{)}^2=4\text{J}$

The final kinetic energy of the system is,

$K.E_f=\frac{1}{2}\times (2+2)\times v^2$

$=\frac{1}{2}\times 4\times (1{)}^2=2\text{J}$

Using equation $\left(1\right)$, we get

$K.E_{loss}=4-2=2\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.