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Physics Question on laws of motion

A gun applies a force F on a bullet which is given by F=(100−0.5105×t)NF = (100 - 0.5 10^5 \times t )N. The bullet emerges out with speed 400 m/s−1400 \,m/s^{-1}. Then find out the impulse exerted till force on bullet becomes zero

A

0.2 N s

B

0.3 N s

C

0.1 N s

D

0.4 N s

Answer

0.1 N s

Explanation

Solution

F = (100 - 0.5 ? 10t)N Given F = 0 100 - 0.5 ? 10t = 0 100 = 0.5 ? 10t T = 2 ? 10 sec I = ∫\intF dt I = ∫02×10−3\displaystyle \int_0^{2\times10^{-3}}(100- 0.5 10 t)dt I=[100t−1052t22]02×10−3=\left[100 t-\frac{10^{5}}{2}\frac{t^{2}}{2}\right]_{_{_{_0}}}^{^{^{2\times10^{-3}}}} I=[100×2×10−3−1054×4×10−6]=\left[100\times2\times10^{-3}-\frac{10^{5}}{4}\times4\times10^{-6}\right] I=[2×10−1−10−1]=\left[2\times10^{-1}-10^{-1}\right] I=10−1=0.1=10^{-1}=0.1 N-s