Question

A $2\text{ }kg$ body moving at a speed of $6\,m{{s}^{-1}}$ collides with another body of $\text{9 }kg$ at rest and sticks together. The loss of kinetic energy would be-
(A). $30J$
(B). $40J$
(C). $33J$
(D). $32J$

Answer 1

Hint: Before collision, both bodies will have different states of motion and after collision, both bodies will move together with equal velocity. Using the law of conservation of momentum, we can calculate the final velocity. After collision, the kinetic energy decreases, so loss of kinetic energy is the difference of initial and final kinetic velocity.

Formulas used:
$p=mv$
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
Given, a body of mass $2\text{ }kg$ is moving with velocity $6\,m{{s}^{-1}}$. Another body of mass$\text{9 }kg$ is at rest.
The formula for momentum,$p$ is,
$p=mv$
Here,$m$ is the mass
$v$is the velocity
Sum of initial momentum of both bodies is-
$\begin{align}
  & {{p}_{i}}={{m}_{1}}{{v}_{1}}+m_{1}^{'}v_{1}^{'} \\
 & \Rightarrow {{p}_{i}}=2\times 6+9\times 0 \\
 & \\
\end{align}$
${{p}_{i}}=12\,kg\,m{{s}^{-1}}$ --- (1)
After collision, both bodies stick together and move with velocity,$V$
The final momentum of the bodies is-
$\begin{align}
  & {{p}_{f}}={{m}_{2}}V+m_{1}^{'}V \\
 & \Rightarrow {{p}_{f}}=2\times V+9\times V \\
\end{align}$
${{p}_{f}}=11V$ ---- (2)
Since the system of bodies is isolated, its momentum is conserved, i.e.
 Initial momentum =final momentum
Therefore, from eq (1) and eq (2), we get,
$12=11V$
$\Rightarrow V=\dfrac{12}{11}\,m{{s}^{-1}}$ --- (3)
We know,
$K=\dfrac{1}{2}m{{v}^{2}}$
Here, $K$ is the kinetic energy of a body
Sum of initial kinetic energies of the bodies will be-
$\begin{align}
  & K={{K}_{1}}+{{K}_{2}} \\
 & K=\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2} \\
 & K=\dfrac{1}{2}\times 2\times {{(6)}^{2}}+0 \\
\end{align}$
 $\Rightarrow K=36J$ --- (4)
The final kinetic energy of the system of bodies will be-
$\begin{align}
  & K'=\dfrac{1}{2}({{m}_{1}}+{{m}_{2}}){{V}^{2}} \\
 & K'=\dfrac{1}{2}(9+2){{\left( \dfrac{12}{11} \right)}^{2}} \\
\end{align}$
 $\Rightarrow K'=6.5J$ --- - (5)
The loss in Kinetic energy is-
 $K-K'$
 Therefore from eq (4) and eq (5), we get,
$\begin{align}
  & \Delta K=36-6.5 \\
 & \Delta K=29.5J\approx 30J \\
\end{align}$

Therefore, the loss in kinetic energy is $30J$. So the correct option is (A).

Note:
Before collision, the two bodies are seen as separate systems. After collision, the two bodies stick together to form one system. In an isolated system the momentum before and after a collision is conserved. Momentum is a vector quantity; it depends on the direction of the motion of bodies.