Question

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.

Answer 1

Mass of bullet (m) = 10 g = 0.01 kg

The initial velocity of a bullet (u) = 150 ms^-1

The final velocity of a bullet (v) = 0

Time (t) = 0.03 s

Acceleration on the bullet (a) =?

Distance penetrated by a bullet (s) =?

Force acting on wooden block (F) =?

(i) Applying, v = u + at

⇒ 0 = 150 m s^-1 + a × 0.3 s

⇒ - a × 0.03 s = 150 m s^-1

or a = `- (150  ms^-1)/(0.03s)`

= - 5000 ms^-1

Applying, s = `ut + 1/2 at^2`

= `150  ms^-1 xx 0.03 s + 1/2 xx (-5000  ms^-2) xx (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

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