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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

AIIMS 2019 · 25 May · Physics · Center of Mass, Momentum and Collision

A gun applies a force F on a bullet which is given by F=(100-0.5 ×...

A gun applies a force FF on a bullet which is given by F=(100−0.5×105t)NF=\left(100-0.5 \times 10^5 t\right) \mathrm{N}. The bullet emerges out with speed 400 m s−1400 \mathrm{~m} \mathrm{~s}^{-1}. Then find out the impulse exerted till force on bullet becomes zero

Options

  1. A0.2 N s0.2 \mathrm{~N} \mathrm{~s}
  2. B0.3 N s0.3 \mathrm{~N} \mathrm{~s}
  3. C0.1 N s0.1 \mathrm{~N} \mathrm{~s}
  4. D0.4Ns0.4 \mathrm{Ns}

Correct answer

C. 0.1 N s0.1 \mathrm{~N} \mathrm{~s}

Step-by-step solution

 (c) : Given, force F=(100−0.5×105t)N100−0.5×105t=0,100=0.5×105t(∵F=0)t=2×10−3sec Impulse, I=∫Fdt=∫02×10−3(100−0.5×105t)dtI=[100t−1052t22]02×10−3I=[100×2×10−3−1054×4×10−6]I=10−1=0.1Ns\begin{aligned} & \text{ (c) : Given, force } F=\left(100-0.5 \times 10^5 t\right) \mathrm{N} \\ & 100-0.5 \times 10^5 t=0,100=0.5 \times 10^5 t(\because F=0) \\ & t=2 \times 10^{-3} \mathrm{sec} \\ & \text{ Impulse, } I=\int F d t=\int_0^{2 \times 10^{-3}}\left(100-0.5 \times 10^5 t\right) d t \\ & I=\left[100 t-\frac{10^5}{2} \frac{t^2}{2}\right]_0^{2 \times 10^{-3}} \\ & I=\left[100 \times 2 \times 10^{-3}-\frac{10^5}{4} \times 4 \times 10^{-6}\right] \\ & I=10^{-1}=0.1 \mathrm{ N} \mathrm{ s}\end{aligned}
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