91Ó°ÊÓ

We have given,

Mass of clay ball=$50\mathrm{g}$

Speed of clay ball =${\mathrm{v}}_0$

Mass of brick =$1\mathrm{kg}$

Speed of brick =$0$

We have to find the speed of the brick after the collision.

According to the law of conservation of momentum, in a inelastic Collison the momentum of the system before and after will be same.

Then,

momentum before collision is given by,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}=\mathrm{momentum}\mathrm{of}\mathrm{clay}+\mathrm{momentum}\mathrm{of}\mathrm{brick} \\ {\mathrm{P}}_{\mathrm{i}}={\mathrm{m}}_{\mathrm{clay}}{\mathrm{v}}_{\mathrm{clay}}+{\mathrm{m}}_{\mathrm{brick}}{\mathrm{v}}_{\mathrm{brick}} \\ {\mathrm{P}}_{\mathrm{i}}=(0.05\mathrm{kg})\left({\mathrm{v}}_0\right)+(1\mathrm{kg})\left(0\right) \\ {\mathrm{P}}_{\mathrm{i}}=0.05{\mathrm{v}}_0\mathrm{kg}.\mathrm{m}/\mathrm{s}......................\left(1\right) \end{gathered}$$

After the collision the clay will stick with the brick, then momentum will be,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{f}}=\mathrm{momentum}\mathrm{of}\mathrm{clay}+\mathrm{momentum}\mathrm{of}\mathrm{brick} \\ {\mathrm{P}}_{\mathrm{f}}=(0.05\mathrm{kg}+1\mathrm{kg})\left({\mathrm{v}}_{\mathrm{f}}\right) \\ {\mathrm{P}}_{\mathrm{f}}=1.05{\mathrm{v}}_1\mathrm{kg}.\mathrm{m}/\mathrm{s}.............\left(2\right) \end{gathered}$$

From the conservation of momentum by equating (1) and (2),

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}} \\ 0.05{\mathrm{v}}_0=1.05{\mathrm{v}}_1 \\ {\mathrm{v}}_1=0.0476{\mathrm{v}}_0\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

mass of clay =$50\mathrm{g}$

mass of brick =$1\mathrm{kg}$

velocity of clay =${\mathrm{v}}_0$

We have to find the percentage of mechanical energy lost in the collision.

We are not able to use here the conservation of energy since the kinetic energy is not conserved in the inelastic collision.

Kinetic energy of the system before the collision is equal to the kinetic energy of the clay ball since the brick is in rest before collision.

then,

$$\begin{gathered} \mathrm{kinetic}\mathrm{energy}=\frac{1}{2}{\mathrm{m}}_{\mathrm{clay}}{\mathrm{v}}_0^2 \\ \mathrm{kinetic}\mathrm{energy}=\frac{1}{2}(0.05\mathrm{kg})\left({\mathrm{v}}_0^2\right) \\ \mathrm{kintic}\mathrm{enregy}=0.025{\mathrm{v}}_0^2..............\left(3\right) \end{gathered}$$

The kinetic energy after the collision is the kinetic energy of the brick and the clay together.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}({\mathrm{m}}_{\mathrm{clay}}+{\mathrm{m}}_{\mathrm{brick}})\left({\mathrm{v}}_{\mathrm{f}}^2\right) \\ {\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}(1.05\mathrm{kg}){(0.0476{\mathrm{v}}_0)}^2 \\ {\mathrm{E}}_{\mathrm{f}}=0.001189{\mathrm{v}}_0^2.............\left(4\right) \end{gathered}$$

Then, the percentage change in the kinetic energy is,

$$\begin{gathered} \%∆=\frac{\mathrm{kinetic}\mathrm{energy}\mathrm{before}\mathrm{collosion}-\mathrm{kinetic}\mathrm{energy}\mathrm{after}\mathrm{colliosin}}{\mathrm{kineic}\mathrm{enrgy}\mathrm{before}\mathrm{colliosin}}×100 \\ \%∆=\frac{0.025{\mathrm{v}}_0^2-0.001189{\mathrm{v}}_0^2}{0.025{\mathrm{v}}_0^2}×100 \\ \%∆=95.24\% \end{gathered}$$