A 5 g bullet moving at 100 m·s\(^{-1}\) strikes a police dog and lodges in his shoulder. The bullet undergoes uniform acceleration and penetrates to a depth of 6 cm. Calculate the time taken for the bullet to stop.

Since the bullet undergoes uniform acceleration, its average speed while inside the dog is 1/2 of (initial speed + final speed) = 1/2(100 + 0) = 50 m/s .
It travels 6 cm at that average speed.
Time = (distance) / (speed) = (6 cm) / (50 m/s) = (0.06 m) / (50 m/s) = 0.0012 sec
(1.2 milliseconds) .

Answered by AL2006 | 2024-06-10

*** V² = U² + 2 a S***
Here V = final velocity = 0 of the bullet as it stopped *** U = initial velocity before the bullet entered the shoulder = 100 m/s***
*** a = ?***
*** S = distance traveled by bullet under the acceleration/deceleration. = 6 cm***
*** = 0.06 meter***
a = (0² - 100²) / 2 * 0.06 = - 83,333 meters/sec² = - 8.333 * 10^4 m/sec² It is negative as it is a deceleration . Now ,let us calculate the time duration T taken by the bullet to stop after entering the shoulder.
V = u + a t Here, we have V = final velocity = 0. u is 100 m/s a is found above. So 0 = 100 - 83333 T => T = 100 / 83333 = 1.2 milliseconds

Answered by Everest2017 | 2024-06-10

The bullet takes approximately 0.0012 seconds, or 1.2 milliseconds, to stop after penetrating 6 cm into the dog's shoulder. This is calculated using the average speed during penetration. The average speed is calculated as half of the initial and final speeds of the bullet.
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Answered by AL2006 | 2024-10-16