Question

A body of mass m at rest explodes into three piece to of which of mass m upon 4 each are thrown off in perpendicular direction with velocities of 3 metre per second 4 metre per second respectively the third piece will be thrown off with a velocity ram

Answer 1

Answer:

Initial mass = M

initial velocity = 0

After explosion, 

m1 = M/4

v1 = 3m/s i (i means along x-direction)

m2 = M/4

v2 = 4m/s j (j means along y-direction)

m3 = M-(M/4 + M/4) = M/2

v3 = V

As there is no external force, 

momentum before explosion = momentum after explosion

⇒M(0) = m1v1 + m2v2 + m3v3

⇒ 0 = M/4 (3)i + M/4 (4)j + M/2(V)

⇒ 0 = M [ (3/4)i + 1j + V/2 ]

⇒ 0 =  (3/4)i + 1j + V/2

⇒ V/2 = -[(3/4)i + 1j]

⇒ V = 2×-[(3/4)i + 1j]

⇒ V = -[ (3/2)i +(2)j ]

⇒ V = -(3/2)i -2j

|V| = √[(3/2)² + 2²] = 2.5m/s

Answer 2

Apply law of conservation of leinear momentum to solve it fast.......

see the attachement.....