JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 20)
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 :
0.3 ms-1
0.1 ms-1
0.7 ms-1
0.4 ms-1
Explanation
Given, resistance offered by the wall
$$ =F=-25 \times 10^{-2} \mathrm{~N} $$
So, deacceleration of bullet,
$$ \begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\\\ (\because m & \left.=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right) \end{aligned} $$
Now, using the equation of motion,
$$ v^2-u^2=2 a s $$
We have,
$$ v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right) $$
$$ \left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right) $$
$$ \begin{array}{ll} \Rightarrow v^2=\frac{1}{2} \\\\ \therefore v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1} \end{array} $$
$$ =F=-25 \times 10^{-2} \mathrm{~N} $$
So, deacceleration of bullet,
$$ \begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\\\ (\because m & \left.=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right) \end{aligned} $$
Now, using the equation of motion,
$$ v^2-u^2=2 a s $$
We have,
$$ v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right) $$
$$ \left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right) $$
$$ \begin{array}{ll} \Rightarrow v^2=\frac{1}{2} \\\\ \therefore v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1} \end{array} $$
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