Verified Solution • 2025–26 Session moderate

QJEEM19Pha353: JEE Main 2019 Physics - Class 11 JEEM19Pha353 Solution: A bullet of mass 20 g has an initial speed of 1 ms –1 , just…

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 :

Multiple Choice Options

  • A 0.3 ms-1
  • B 0.1 ms-1
  • C 0.7 ms-1
  • D 0.4 ms-1

Step-by-Step Verified Solution

Given, resistance offered by the wall

=F=−25×10−2 N=F=-25 \times 10^{-2} \mathrm{~N}

So, deacceleration of bullet,

a=Fm=−2.5×10−220×10−3amp;=−54 ms−2(∵mamp;=20 g=20×10−3 kg)\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,

v2−u2=2asv^2-u^2=2 a s

We have,

v2=1+2(−54)(20×10−2)v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right)

(∵u=1 ms−1 and s=20 cm=20×10−2 m)\left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right)

⇒v2=12∴v=12≈0.7 ms−1\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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