A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular directions with speed v. What will be the speed of the third (lighter) fragment?
Answers
After the explosion, let the masses of the three fragment be 2m, 2m, and m respectively.
Applying the law of conservation of momentum,
Initial Momentum = Final momentum.
Now, after the explosion, 2m masses moves along perpendicular direction. Let assume this to be in x-direction and one in y-direction.
If then will move in two perpendicular direction, then other mass should move in such an way that its momentum is just opposite in direction equal to both.
Refer to the Attachment,
In X-direction.
Initial Momentum = Final momentum.
∴ 0 = mucosA - 2mv
∴ mucosA = 2mv
∴ ucosA = 2v -----(1).
Similarly, In y-direction,
0 = musinA - 2mv
muSinA = 2mv
∴ uSinA = 2v ---(2).
(2) ÷ (1).
tanA = 1
∴ A = 45°
Thus, uSinA = 2v
∴ u = 2√2v
''ux = 2v and uy = 2v''
Hope it helps.
