A police dog gets caught in a shootout. A 5 g bullet moving at 100 m·s-1 strikes the dog and lodges in his shoulder. Fortunately the dog survives. The bullet undergoes uniform acceleration and penetrates his shoulder to a depth of 6 cm. Calculate the time taken for the bullet to stop

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

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).