Question

A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 900 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.

A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 90 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.

Answer 1

Given: Mass of bullet, m₁ = 10 g = 10 × 10⁻³ kg

The initial speed of the bullet, u₁ = 1.5 m/s

Mass of plank, m₂ = 900 g = 900 × 10⁻³ kg

Initial speed of plank, u₂ = 0 m/s

To find: velocity, v = ?

Formula: From the law of conservation of momentum,

`P_"Initial" = P_"Final"`

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Since the bullet gets embedded in the plank and both move with the same speed, v₁ = v₂ = v

So, we can rewrite the equation as

m₁u₁ + m₂u₂ = (m₁+ m₂)v

10 × 10⁻³ kg × 1.5 m/s + 900 × 10⁻³ kg × 0 m/s = (10 × 10⁻³ kg + 900 × 10⁻³ kg)v

= 1.5 × 10⁻² + 0

= 910 × 10⁻³ kg × v

v = `(1.5 xx 10^(-2))/(910 xx 10^(-3))`

v = `(1.5 xx 10^(-2))/(91 xx 10^1 xx 10^(-3))`

v = `(1.5 xx 10^cancel(-2))/(91 xx 10^cancel(-2))`

v = `(1.5)/(91)`

v = 0.0165 m/s = 1.65 cm/s

Answer 2

According to Question:

m₁ = 10g = 10 × 10⁻³ kg

u₁ = 1.5 m/s 

m₂ = 90g = 90 × 10⁻³ kg

u₂ = 0 m/s,

v₁ = v₂ = v = ?

Based on the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

But u₂ = 0 m/s and v₁ = v₂ = v

m₁u₁ = (m₁ + m₂)v

v = `(m_1u_1)/("m"_1 + m_2)` 

= `(10 xx 10^-3 kg xx 1.5 m "/"s)/(10 xx 10^-3 kg + 90 xx 10^-3 "kg")`

=` (10 xx 10^-3 kg xx 1.5 m"/"s)/(10^-3(10 + 90)kg)`

= `(10 xx 1.5)/100` m/s

= 0.15 m/s