Laws Of Motion Ques 7
- A bullet of mass $20 \mathrm{~g}$ has an initial speed of $1 \mathrm{~ms}^{-1}$, just before it starts penetrating a mud wall of thickness $20 \mathrm{~cm}$. If the wall offers a mean resistance of $2.5 \times 10^{-2} \mathrm{~N}$, the speed of the bullet after emerging from the other side of the wall is close to
(Main 2009, 10 April II)
(a) $0.3 \mathrm{~ms}^{-1}$
(b) $0.4 \mathrm{~ms}^{-1}$
(c) $0.1 \mathrm{~ms}^{-1}$
(d) $0.7 \mathrm{~ms}^{-1}$
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Answer:
Correct Answer: 7.(d)
Solution:
Formula:
- Given, resistance offered by the wall
$$ =F=-2.5 \times 10^{-2} \mathrm{~N} $$
So, deceleration of bullet,
$$ \begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}}= & -\frac{5}{4} \mathrm{~ms}^{-2} \\ & \left(\because m=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right) \end{aligned} $$
Now, using the equation of motion,
$$ v^{2}-u^{2}=2 a s $$
We have,
$$ \begin{aligned} & v^{2}=1+2-\frac{5}{4}\left(20 \times 10^{-2}\right) \\ & \left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right) \\ \Rightarrow \quad & v^{2}=\frac{1}{2} \text{ m}^2/\text{s}^2 \\ \therefore \quad v & =\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1} \end{aligned} $$
BETA
