Question

A bullet is fired through a fixed $10cm$ thick board, in such a manner that the bullet's line of motion is perpendicular to the face of the board. If the initial speed of the bullet is $400m{s}^{-1}$ and it emerges from other side of the board with a speed of $300m{s}^{-1}$ , then

(a) find the deceleration of bullet [assume constant] as it passes the board.

(b) find the total time the bullet is in contact with the board.

Answer 1

Step 1: Given Data

The initial velocity of the bullet $u=400m{s}^{-1}$

The final velocity of the bullet $v=300m{s}^{-1}$

The thickness of the board $s=10cm=0.1m$

Step 2: Determine the deceleration of the bullet as it passes the board

As we know the third equation of motion is ${\mathit{v}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{u}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{a}\mathit{s}$

On putting the given data in the equation we get,

$\Rightarrow {\left(300\right)}^2={\left(400\right)}^2-2a(0.1)$

On further solving we get,

$\Rightarrow a=\frac{160000-90000}{0.2}$

$\Rightarrow a=\frac{70000}{0.2}=3.5\times {10}^{5}m{s}^{-2}$

Step 3: Determine the total time for which the bullet is in contact with the board

As we know the first equation of motion is $\mathit{v}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathbf{}\mathbf{+}\mathbf{}\mathit{a}\mathit{t}$

On putting the given data in the above equation we get,

$\Rightarrow t=\frac{300-400}{-3.5\times {10}^5}$

On further solving we get,

$\Rightarrow t=2.86\times {10}^{-4}s$

Final Answer:

(a) The deceleration of the bullet as it passes the board is $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{2}}$

(b) The total time the bullet is in contact with the board is $\mathbf{2}\mathbf{.}\mathbf{86}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{}}\mathit{s}$ .