Question

A bullet of mass 20 g travelling horizontally with a velocity of 100 m$s^{-1}$ strikes a wooden block and comes to rest in 0.05 s. Calculate the magnitude of force in N exerted by the wooden block on the bullet.

Answer 1

Mass of bullet (m)= 20 g = 0.02 kg

Initial velocity of bullet (u) =100 m$s^{-1}$

Final velocity of bullet (V) = 0

Time (t)= 0.05 seconds

Acceleration on the bullet (a)?

Force acting on wooden block (F) = ?

Distance penetrated by bullet (s) = ?

We know;

Step 1$:$ Calculate the acceleration of the v = u + at

0= 100+ a ×0.05

– a × 0.05 = 100

a = -2000 m$s^{-2}$

Step 2$:$ Calculate the distance of penetration using s = ut + $\frac{1}{2}$ a$t^2$

= 100 × 0.05 + 0.5 x (–2000) × ${.05}^2$

= 7.5 m

We know, F = ma

Force acting on bullet (F)

= 0.02 × (– 2000) = -40N

Negative sign denotes that wooden block exerts force in the direction, opposite to the direction of motion of the bullet.