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

Explanation

Mass of the bullet `m_(1) = 20 g=0.02 kg`.
Mass of the pendulum `m_(2)=5 kg`
Centre of mass of pendulum `m_(2)=5 kg`
Centre of mass of pendulum rises to a
height = h 10 cm = 0.1 m
Speed of the bullet `=u_(1)`
Pendulum is at rest `therefore u_(2)=0`
Common velocity of the bullet and the pendulum after the bullet is embeded into the object = v
`therefore v = (m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))`
`v=(0.02xxu_(1))/(0.02+5)=(0.02 u_(1))/(5.02) " "` ....(1)
From II equation of motion
`v = sqrt(2gh)=sqrt(2xx9.8xx0.1)=sqrt(1.96)=1.4 ms^(-1)`.
Substitute the value of v in equation (1)
`1.4=(0.02 u_(1))/(5.02)`
`therefore u_(1)=(5.02xx1.4)/(0.02)`
`u_(1)=351.4 ms^(-1)`.