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

Options

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

Answer

C

Explanation

Ignoring `g` compared to initial velocity, the bullet and block will meet after `t=(h)/(u)=(100)/(100)=1sec`
Durinig this time, distance travelled by the block
`s_(1)=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m`
Distance travelled by the bullet,
`s_(2)=100-s_(1)=100-5=95m`
velocity of bullet just before collision,
`u_(1)=100-10xx1=90 m//s,` upwards
velocity of block just before collision,
`u_(2)=10xx1=10 m//s, ` downwards
According to law of conservation of linear momentum , `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V`
`0.01(-10+90)=0.02V`
`V=40 m//s`
Maximum height risen by the block
`=(V^(2))/(2g)=(40xx40)/(2xx10)=80m`
Height reached above the top of the tower
`=80-s_(1)=80-5`
`=75m`