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

velocity, v = ?

From the law of conservation of momentum,

`"P"_"Initial" = "P"_"Final"`

`"m"_1"u"_1 + "m"_2"u"_2 = "m"_1"v"_1 + "m"_2"v"_2`

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

`"m"_1"u"_1 + "m"_2"u"_2 = ("m"_1+ "m"_2)"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 xx 10^-2 + 0 = 910 xx 10^-3 "kg" xx "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:

`"m"_1 = 10"g" = 10 xx 10^-3 "kg", "u"_1 = 1.5 "m" "/""s", "m"_2 = 90"g" = 90 xx 10^-3 "kg", "u"_2 = 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"_1"u"_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