At a certain height a shell at rest explodes into two equal fragments. One of the fragments receives a
horizontal velocity u. The time interval after which, the velocity vectors will be inclined at 120° to each other is pls help
Answers
Answer:
The answer will be u/10√3
Explanation:
According to the problem two fragments of shell are created after the exploding which is done an external force.
As here is no external force therefore the momentum will be conserved.
Therefore from momentum conservation we can say that
p= m/2 u+ m/2 v [ where p = momentum]
as p = 0
v= -u
this says that both fragments are opposite to each other with same speed.
Let after t time the fragments will 120 degree to each other.
Now as the one fragment goes horizontally
therefore, the vector will be ,a= ui-gt
= ui- 10t
and the another one will be b= -ui-10t
now the angel between them is 120 degree
a.b = abcosθ
cosθ = a.b/ab
cos 120 degree= a.b/ab
=> -1/2= -u^2+100t^2/100t^2+u^2
after solving
t= u/10√3
Answer:
The answer will be u/10√3
Explanation:
According to the problem two fragments of shell are created after the exploding which is done an external force.
As here is no external force therefore the momentum will be conserved.
Therefore from momentum conservation we can say that
p= m/2 u+ m/2 v [ where p = momentum]
as p = 0
v= -u
this says that both fragments are opposite to each other with same speed.
Let after t time the fragments will 120 degree to each other.
Now as the one fragment goes horizontally
therefore, the vector will be ,a= ui-gt
= ui- 10t
and the another one will be b= -ui-10t
now the angel between them is 120 degree
a.b = abcosθ
cosθ = a.b/ab
cos120 degree= a.b/ab
=> -1/2= -u^2+100t^2/100t^2+u^2
after solving
t= u/10√3