Question

A bullet fired with a speed of 50 $\frac{m}{s}$ is stopped at a depth of 10 cm by a wall .What is the retardation offered by the wall ?

Answer 1

Step 1:

Given that

Initial Speed of the bullet fired(u) = 50 $\frac{m}{s}$.

Final speed ( v) = 0.

Distance (S) =10 cm = 0.1 m

Step 2: Formula .

Retardation = $v^2-u^2=2as$ or $a=\frac{v^2-u^2}{2s}.$ ( u is initial speed , v is final speed , a is acceleration and s is distance ) .

Step 3 :Calculation

Retardation = $v^2-u^2=2as$ or $a=\frac{v^2-u^2}{2s}.$

Retardation = $a=\frac{v^2-u^2}{2s}.$

$$\begin{gathered} a=\frac{\hspace{0.17em}(0^2-{50}^2)}{(2\times 0.1)}. \\ a=\frac{-2500}{0.2}=-12500\frac{m}{s^2}. \end{gathered}$$

Step 4:

Retardation offered by the wall is = $12500\frac{m}{s^2}.$