An object of mass 5 kg is projecte with a velocity of `20ms^(-1)` at an angle of `60^(@)` to the horizontal. At the highest point of its path , the projectile explodes and breaks up into two fragments of masses 1kg and 4 kg. The fragments separate horizontally after the explosion, which releases internal energy such that `K.E.` of the system at the highest point is doubled. Calculate the separation betweent the two fragments when they reach the ground.

Options

A. 52.25m
B. 44.25m
C. 65.32m
D. 78.76m

Answer

B

Explanation

Given , `m=5kg, v= 20 ms^(-1), theta= 60^(@)`
Vertical component of velocity, `v_(y)= v sin 60^(@)`
`=20 xx (sqrt3)/(2) =10 sqrt3 ms^(-1)`
Time taken to reach the highest point = Time taken to reach the ground from highest point.
`t= (v sin theta)/(g)= (v_(y))/(g)= (10 sqrt3)/(9.8) = 1.77s`
If the highest point, m splits up into two parts of masses `m_(1)=1kg and m_(2)=4kg`
If their velocities `v_(1) and v_(2)` respectively, then applying the principle of conservation of linear momentum, we get
`m_(1)v_(1) +m_(2)v_(2)= mv cos theta`
`v_(1)+ v_(2)=5 xx 20 xx (1)/(2) [because theta= 60^(@)]`
`v_(1)+ 4v_(2)=5 xx 10=50` ...(i) ,br> Initital KE `=(1)/(2) m (v cos theta)^(2)`
`=(1)/(2) xx 5 xx (10)^(2)= 250J`
Final KE= 2 (initial KE)= `2 xx 250= 500J`
`therefore (1)/(2) m_(1)v_(1)^(2) + (1)/(2) m_(2) v_(2)^(2)= 500`
or `(1)/(2) xx 1 xx v_(1)^(2) + (1)/(2) xx 4 xx v_(2)^(2)= 500`
or `v_(1)^(2)+ 4v_(2)^(2)=1000` ...(ii)
Solving Eqs (i) and (ii), we get
`v_(1)=30 m//s, v_(2)= 5m//s`
Hence, the separation between teh two fragments
`=(v_(1)-v_(2))xx t= (30-5)xx 1.77m= 44.25m`