A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^-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.

Given,

Mass of the bullet (m) = 10 g

Convert g into kg

1000 g = 1 kg

So, 10 g = $\dfrac{10}{1000}$ = 0.01 kg

Initial velocity of the bullet (u) = 150 ms^-1

Terminal velocity of the bullet (v) = 0

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{0-150}{0.03} = \dfrac{150}{0.03} =\dfrac{-15000}{3}$ = -5000 ms^-2

∴ Acceleration = -5000 ms^-2.

According to the third equation of motion,

2as = v² - u²

or

a = $\dfrac{\text{v}^2 - \text{u}^2 }{2\text{s}}$

Substituting we get,

s = $\dfrac{0 - (150)^2}{2 \times -5000} = \dfrac{-22500}{-10000} =$ = 2.25 m.

Hence, the total distance travelled is 2.25 m

Now, as Force = Mass x Acceleration

Substituting we get,

F = 0.01 × -5000 = -50 N

Negative sign indicates direction of recoil of bullet is opposite to the direction of bullet.

Hence, magnitude of the force = 50 N