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  4. A gun applies a force F on a bullet which is given by F = (100 - 0.5 105 × t )N. The bullet emerges out with speed 400 m/s-1. Then find out the impulse exerted till force on bullet becomes zero

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Q. A gun applies a force F on a bullet which is given by $F = (100 - 0.5 10^5 \times t )N$. The bullet emerges out with speed $400 \,m/s^{-1}$. Then find out the impulse exerted till force on bullet becomes zero

AIIMSAIIMS 2019Laws of Motion

Solution:

F = (100 - 0.5 × 105t)N
Given F = 0
100 - 0.5 × 105t = 0
100 = 0.5 × 105t
T = 2 × 10-3 sec
I = $\int$F dt
I = $\displaystyle \int_0^{2\times10^{-3}}$(100- 0.5 105 t)dt
I$=\left[100 t-\frac{10^{5}}{2}\frac{t^{2}}{2}\right]_{_{_{_0}}}^{^{^{2\times10^{-3}}}}$
I$=\left[100\times2\times10^{-3}-\frac{10^{5}}{4}\times4\times10^{-6}\right]$
I$=\left[2\times10^{-1}-10^{-1}\right]$
I$=10^{-1}=0.1$ N-s