A wooden block of mass `10 g` is dropped from the top of a tower `100 m` high. Simultaneously, a bullet of mass `10 g` is fired from the foot of the tower vertically upwards with a velocity of `100 m//s`. If the bullet is embedded in it, how high will the block rise above the top of tower before it starts falling? `(g=10m//s^2)`

Options

A. `75m`
B. `85m`
C. `80m`
D. `10m`

Answer

A

Explanation

The bullet and block will meet after tme `t=h/(u_(rel))=100/100=1`
During this time, distance travelled by the block,
`s_(1)=1/2"gt"^(2)=1/2xx10xx1^(2)=5m`
Distance travelled by the bullet,
`s_(2)=100-s_(1)=96m`
Velocity of the bullet before collision
`u_(2)=u-"gt"=100-10xx1=90m//s`
Velocity of the block before collision.
`u_(1)"gt"=10m//s`
Let `V` be the combined velocity after collision.
According to the law of conservation of momentum.
`m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V`
(Velocity in upward direction in considered positive)
Solving we get `V=40m//s`
Maximum height risen by the block `=(V^(2))/(2g)=80m`
Height reached above the top of the tower is
`80-s_(1)=80-5=75m`