A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Mass of bullet (m) = 10 g = 0.01 kg Initial velocity of bullet (u) = 150 m s-1
Final velocity of bullet (v) = 0
Time (t) = 0.03 s
Acceleration on bullet (a) = ?
Distance penetrated by bullet (s) = ?
Force acting on wooden block (F) = ?
(i) Applying, v = u + at
=> 0 = 150 ms-1 + a x 0.3 s
=> -a x 0.03 s = 150 ms-1
or a = \(-\frac{150 ms^{-1}}{0.03s}\) = -5000 ms-1
Applying, s = ut + \(\frac{1}{2}at^2\)
= 150ms-1 x 0.03s + \(\frac{1}{2}\) x (-5000 ms-2) x (0.03 s)2
= 4.5 m - 2.25 m = 2.25 m
(ii) Applying, F = ma Force acting on bullet (F) = 0.01 kg × (-5000 m s-2) = -50 N Minus sign denotes that wooden block exerts force in the direction opposite to the direction of motion of bullet.