91Ó°ÊÓ

Mass of a bullet, $m=30g=0.030kg$

Distance, $d=12cm=0.12m$

A horizontal initial velocity, $v_i=500m/s$

Final velocity, $v_f=0m/s$

As the bullet is moving in the horizontal direction, there will be no change in the potential energy of the bullet. Hence, you can neglect it. Using the change in kinetic energy of the bullet, you can find the change in the bullet’s mechanical energy.

As known that work done by the force is equal to the change in mechanical energy. Using this, you equate this to the formula for work to find the required force.

Formulae:

The change in kinetic energy is,

$∆KE=K{E}_{final}-K{E}_{initial}$

Here, $K{E}_{initial}andK{E}_{final}$and are the initial and final kinetic energy respectively.

The kinetic energy is defined by,

$KE=\frac{1}{2}m{v}^2$

Write the work done equations as below.

$W=F∆\mathrm{d}$ ….. (1)

And

$W=\left|∆E_{mech}\right|$ ….. (2)

Here, W is the work done, F is the force, $∆d$ is the change in distance, and $∆E_{mech}$ is the change in mechanical energy.

Calculate the initial kinetic energy as below.

$$\begin{gathered} K{E}_{initial}=\frac{1}{2}m{v}_i^2 \\ =\frac{1}{2}×0.03\mathrm{kg}×{\left(500\mathrm{m}/\mathrm{s}\right)}^2 \\ =3750\mathrm{J} \\ =3.8×{10}^3\mathrm{J} \end{gathered}$$

Determine the final kinetic energy as below.

$$\begin{gathered} K{E}_{final}=\frac{1}{2}m{v}_f^2 \\ =\frac{1}{2}×0.03kg×{(0m/s)}^2=0J \end{gathered}$$

Therefore, the change in kinetic energy is,

$$\begin{gathered} ∆KE=K{E}_{final}-K{E}_{initial} \\ =0-3.8×{10}^3\mathrm{J} \\ =-3.8×{10}^3\mathrm{J} \end{gathered}$$

As the ball travels in the horizontal direction, there is no change in potential difference.

$$\begin{gathered} ∆E_{mech}=∆KE \\ =-3.8×{10}^{3}J \end{gathered}$$

Hence, the change in the bullet’s mechanical energy will be $-3.8×{10}^3\mathrm{J}$.

From equation (1) and (2), you can write the following equation.

$$\begin{gathered} F∆d=|∆E_{mech}| \\ F=\frac{\left|∆E_{mech}\right|}{d} \\ =\frac{3.8×{10}^{8}J}{0.12m} \\ =3.1×{10}^{4}N \end{gathered}$$

Hence, the magnitude of the average force from the wall to stop the bullet will be $3.1×{10}^{4}N$.