Force and Laws of Motion — Question 13

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