Question 127: Laws of Motion

Question

A bullet of mass 20 g has an initial speed of 1 ms–1
, just before it starts penetrating a mud wall of thickness
20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the
other side of the wall is close to :

Accepted Answer

Given, resistance offered by the wall
$$
=F=-25 	imes 10^{-2} \mathrm{~N}
$$
So, deacceleration of bullet,
$$
\begin{aligned}
a=\frac{F}{m}=\frac{-2.5 	imes 10^{-2}}{20 	imes 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\
(\because m & \left.=20 \mathrm{~g}=20 	imes 10^{-3} \mathrm{~kg}ight)
\end{aligned}
$$
Now, using the equation of motion,
$$
v^2-u^2=2 a s
$$
We have,
$$
v^2=1+2\left(-\frac{5}{4}ight)\left(20 	imes 10^{-2}ight)
$$
$$
\left(\because u=1 \mathrm{~ms}^{-1} 	ext { and } s=20 \mathrm{~cm}=20 	imes 10^{-2} \mathrm{~m}ight)
$$
$$
\begin{array}{ll}
\Rightarrow  v^2=\frac{1}{2} \\
	herefore  v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1}
\end{array}
$$

Answer

0.3 ms-1

Answer

0.1 ms-1

Answer

0.7 ms-1

Answer

0.4 ms-1

Laws of Motion

A bullet of mass 20 g has an initial speed of 1 ms–1 , just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the other side of the wall is close to :

Choose the correct answer: