Question

A 3g bullet travelling at 300m/s hits and embeds itself in a 1-kg wooden block resting on a frictionless surface.
a) how far will it slide after being hit by a bullet that exerts 3N friction force

Answer 1

Given :

mass of bullet , m = 3 g = 0.003 kg

velocity of bullet , v = 300 m/s

mass of wooden block , M = 1 kg

frictional force exerted by bullet = 3 N

To find :

Distance to which the wooden block will slide after being hit by the bullet.

Solution :

Let combined velocity of block+bullet system is u .

Let s to be distance to which the bullet+block system slide .

by , conservation of linear momentum ,

    0.003 * 300 = ( 1+0.003) * u

=> u = 0.897 m/s

acceleration of the block+bullet system = 3 N / 1.003 kg

                                                                   = 2.99 m/$s^{2}$ (retarding in nature)

by 3rd equation of motion , where final velocity is 0 (zero) ,

     0 = ( 0.897 * 0.897 ) - ( 2 * 2.99 * s )

=> s = 0.134 m

Distance to which the wooden block will slide after being hit by the bullet is 0.134 m .

Answer 2

The box slides 0.135 m after being hit by a bullet that exerts 3 N friction force.

Given:

Mass of the bullet = m₁ = 3 g = 0.003 kg

Speed of the bullet = v₁ = 300 m/s

Mass of the wooden block = m₂ = 1 kg

Speed of the block = v₂ = 0 m/s (after it hits)

Frictional force = 3 N

To Find:

The distance box slides = ?

Calculation:

Using law of conservation of momentum, we have:

(m₁ × v₁) + (m₂ × v₂) = (m₁ + m₂) × u

On substituting the values, we get,

(0.003 × 300) + (1 × 0) = (0.003 + 1) × u

0.9 = 1.003 × u

u = 0.9/1.003

∴ u = 0.897 m/s

The force is given by the formula:

F = ma

Now, on using equation of motion.

v² = u² + 2as

On substituting the values, we get,

0 = (0.897)² + 2as

a = -0.805/(2s)

a = -0.805/(2s)

Negative sign shows the box is under deceleration.

The force formula becomes,

3 = (1.003) × 0.805/(2s)

s = (1.003) × 0.805/6

∴ s = 0.135 m