Question

A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of the pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

Answer 1

Given:

m₁ = 20 g = 20 × 10^-3 kg; m₂ = 5 kg; s = 10 × 10^-2 m.

Let the speed of the bullet be v. The common velocity of bullet and pendulum bob is V. According to the law of conservation of linear momentum.

V = `(m_1v)/((m_1 + m_2)) = (20 xx 10^-3v)/(5 + 20 xx 10^-3) = 0.02/5.02 v = 0.004  v`

The bob with bullet goes up with a deceleration of g = 9.8 ms^-2. Bob and the bullet come to rest at a height of 10 × 10^-2 m.

from III rd equation of motion

`v^2 = u^2 + 2as` here

`v^2 - 2gs = 0`

`v^2 = 2gs`

`(0.004  v)^2 = 2 xx 9.8 xx 10 xx 10^-2`

`v^2 = (2 xx 9.8 xx 10 xx 10^-2)/(0.004)^2`

v = 350 ms^-1.