A projectile is moving at `20ms^(-1)` at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at `30ms^(-1)` with respect to the gound. Then the other part will move it

Options

A. `20ms^(-1)`
B. `10sqrt(31)ms^(-1)`
C. `50ms^(-1)`
D. `30ms^(-1)`

Answer

C

Explanation

Let m kg be the mass of the projectile.
`therefore` Initial momentum of the projectile
`=mxx20hatikg-m//s`.
After the internal explosion the mass of each part will be m/2.
Momentum of the first part=`(m)/(2)xx30hatjkg-m//s`
Let `vecv` be the velocity of the second part.
Then from the principle of conservation of linear momentum, `mxx20hati=(m)/(2)xx30hatj+(m)/(2)vecv`
`therefore vecv=40hati-30hatj`
`therefore` Magnitude of `|vecv|=sqrt(40^(2)+(-30)^(2))=50m//s`.