First slide

Second law

Question

A bullet is fired from a gun. The force on the bullet is given by F = 600 – (2 × 10⁵t), where F is in newton and 't' is in second. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is

Moderate

Solution

F = 600 – (2 × 10⁵t) = 0
600 = 2 × 10⁵t

$\large t = \frac{{300}}{{{{10}^5}}} = 3 \times {10^{ - 3}}\sec$
Impulse I =

$\large \int\limits_0^p {dp} = \int\limits_0^t {Fdt} ;\;P = \int\limits_0^t {600\,\,dt - 2 \times {{10}^5}} \int\limits_0^t {t\,d\,t}$

$\large = [600\,t]_0^{3 \times {{10}^{ - 3}}} - \cancel{2} \times {10^{ + 5}}\left[ {\frac{{{t^2}}}{{\cancel{2}}}} \right]_0^{3 \times {{10}^{ - 3}}}$

$\large = 1800 \times {10^{ - 3}} - 9 \times {10^{ - 6}} \times {10^5}$

$\large = 1.8 - 0.9 = 0.9\,\,NS$

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