A body of mass 50 kg is projected vertically upwards with a velocity of 100 `m cdot s^(-1)`. After 5 s it splits up into two parts due to an explosion. One part of mass 20 kg moves vertically upwards with a velocity of 150 `m cdot s^(-1)`.Find the velocity of the second body. Find the sum of momenta of the two parts 3 s after the explosion and show that if there was no explosion. the momentum of the body would have been constant.

Explanation

Let the velocity of the projectile 5 s after the projection and just before the explosion be v.
`therefore" " v = 100 - 9.8 xx 5 = 51 m cdot s^(-1)`
Let the velocity of the second part after explosion be `v_(1)`. Appliying the law of conservation of linear momentum along the direction of projection.
` 50 xx 51 = 20 xx 150 + 30 xx v_(1)`
or, `" " v_(1) = (50 xx 51 xx - 3000)/(30) = -15 "m"cdot s^(-1)` (downwards)
Let the velocity of the 20 kg mass, prdouced due to the explosion. 3 seconds after the explosion be v..
`therefore" "v. " " = 150 - 9.8 xx 3 `
= 150 - 29.4 = 120.6`" " m cdot s^(-1)` (upwards)
If v.. is the velocity of the 30 kg mass 3 seconds after explosion, then
v.. = 15 + 9.8`xx` 3
= 15 + 29.4 = 44.4 m`cdot s^(-1)` (downwards)
Thus the total momentum 3 seconds after the explosion
= 20 `xx 120.6 - 30 xx 44.4 = 1080 kg cdot m cdot s^(-1)`
In the case of no explosion, the velocity after 8 seconds of projection would have been
`v_(2) = 100 - 9.8 xx 8 = 100 - 78.4 = 21.6 m cdot s^(-1)`
Hence, its momentum after 8 s would have been `50 xx 21.6 = 1080 "kg" cdot m cdot s^(-1)`.