Question

A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block & comes to rest in 0.03 seconds. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of Force exrerted by wooden block on bullet

Answer 50 newtons ​

Answer 1

A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s.

(mass of bullet is 10 g and initial velocity of the bullet is 150 m/s).

Also given that, the bullet strikes a stationary wooden block & comes to rest in 0.03 seconds. As, the bullet comes to rest. So the final velocity of the bullet is 0 m/s.

Using the First Equation Of Motion,

v = u + at

Substitute the values,

0 = 150 + a(0.03)

-150 = 0.03a

-5000 = a

So, the acceleration of the bullet is -5000 m/s².

Now,

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the known values,

(0)² - (150)² = 2(-5000)(s)

- 22500 = -10000s

225 = 100s

2.25 = s

So, the distance covered by the bullet is 2.25 m.

Also,

Force = mass × acceleration

(mass = 10g = 10/1000 = 0.01 kg)

Force = 0.01 × (5000)

Force = 50 N

Therefore, the Force exrerted by wooden block on bullet is 50N.

Answer 2

Solution :

First of all calculate the acceleration of the bullet

• Initial Velocity (u) = 150 m/s

• Final Velocity (v) = 0

• Time (t) = 0.03 s

From 1st equation of motion :

★ v = u + at

→ 0 = 150 + a × 0.03

→0 = 150 + 0.03 a

→ 0.03 a = - 150

→ a = $\sf\dfrac{-150}{0.03}$

→ a = -5000 m/s²

Now, Let's calculate the Distance of penetration of bullet :

From 3rd equation of motion :

★ v² = u² + 2as

→ (0)² = (150)² + 2(-5000)(s)

→ 0 = 22500 - 10000 × s

→ 10000 s = 22500

→ s = $\sf\dfrac{22500}{10000}$

→ s = 2.25 m

Thus, Distance of penetration of bullet into the block of wood will be 2.25 metres

Let's calculate the magnitude of force

★ Force = Mass × Acceleration

→ Force = $\sf\dfrac{10}{1000}$ kg × -5000 m/s²

→ Force = 50 N

$\therefore$ Magnitude of force exerted by the wooden block on the bullet is 50 newtons