A cannon ball is fired with a velocity v such that it makes an angle θ with the horizontal. At the highest point,the cannon ball splits into two parts of equal masses. One of the parts retraces the initial path of the ball. What will be the speed of the second part immediately after the explosion ?
-
UP 0 DOWN 0 0 3
3 Answers
263At the highest point no vertical velocity exists.Also there is no net force in the horizontal direction.Hence the linear momentum of the system remains conserved along the horizontal direction.As one part retraces the path so it's horizontal velocity just after collision is vcosθ.
Now conserving linear momentum we get,
m*vcosθ= -(m/2)*vcosθ + (m/2)*u
or, u=3vcosθ.
Himanshu Giria y is there a -sign in the first term of RHS
Upvote·0· Reply ·2014-06-19 18:11:43- ARKADYUTI BANDYOPADHYAY OK,thanks.
1161@ARKADYUTI BANDYOPADHYAY: If you got the logic and able to solve this question, then don't forget to post the complete solution from your side.
263Initially the whole ball moves right.finally one part has horizontal velocity towards left after collision.
- Himanshu Giria o ys i actually missed put the word initial in the qn ///