Laws of Motion 1 Question 1
1. 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 $2.5 \times 10^{-2} N$, the speed of the bullet after emerging from the other side of the wall is close to
(Main 2009, 10 April II)
(a) $0.3 ms^{-1}$
(b) $0.4 ms^{-1}$
(c) $0.1 ms^{-1}$
(d) $0.7 ms^{-1}$
Assertion and Reason
Mark your answer as
(a) If Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
(b) If Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I
(c) If Statement I is true; Statement II is false
(d) If Statement I is false; Statement II is true
Show Answer
Answer:
Correct Answer: 1. (d)
Solution:
- Given, resistance offered by the wall
$$ =F=-2.5 \times 10^{-2} N $$
So, deacceleration of bullet,
$$ \begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}}= & -\frac{5}{4} ms^{-2} \\ & \left(\because m=20 g=20 \times 10^{-3} kg\right) \end{aligned} $$
Now, using the equation of motion,
We have,
$$ \begin{aligned} & v^{2}=1+2-\frac{5}{4}\left(20 \times 10^{-2}\right) \\ & \left(\because u=1 ms^{-1} \text { and } s=20 cm=20 \times 10^{-2} m\right) \\ \Rightarrow \quad & v^{2}=\frac{1}{2} \\ \therefore \quad v & =\frac{1}{\sqrt{2}} \approx 0.7 ms^{-1} \end{aligned} $$
$$ v^{2}-u^{2}=2 a s $$
