A shell is fired from a cannon with a velocity `20 m//s` at an angle `60^(@)` with the horizontal. At the highest point it explodes into three pieces of equal masses . One of the pieces retraces its path to the cannon , the second piece moves vertically up with speed `40 m//s` , the third piece will move with velocity (immediately after the explosion)

Options

A. `20 m//s` in a horizontal direction
B. `40 m//s` at an angle `45^(@)` with horizontal in an upward direction
C. `40sqrt(2) m//s` at an angle `45^(@)` with horizontal in an upward direction
D. `40sqrt(2) m//s` at an angle `45^(@)` with horizontal in the downward direction

Answer

D

Explanation

Velocity of shell at the highest point ` = 20 cos 60^(@) = 10 m//s`
Momentum conservation at the highest point
`x : 3m xx 10 = -m xx 10 + m v cos alpha rArr v sin alpha = - 40`
Third piece :
` v = 40 sqrt(2) m//s`
`alpha = 45^(@)`