Question

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

Answer 1

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

Answer 2

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) .