Question

A body of mass M at rest explodes in three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3m/s and 4m/s respectively. The third will thrown off with a velocity of ?

Answer 1

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

Momentum is a vector.
Initial momentum of M is zero.  So resultant of vector sum of all three momenta of 3 pieces is zero. Let us find the resultant of the momenta of two masses M/4.
  
  Momenta are 3 M/4  and  4M/4 = M.  The angle between them is 90⁰.

$R = \sqrt{(3M/4)^2 + M^2 + 2 (3M/4)M\ Cos 90}=1.25M\\ \\Tan \theta = 3/4$
 So θ = 36.87⁰.

So the momentum of the mass M/2 = 1.25 M.
  
    Its velocity = 1.25 M / (M/2)  = 2.5 m/sec
    Its direction is 126.87⁰ from direction of M/4 having velocity 3 m/s  and
                    143.13⁰ from the direction of M/4 having velocity 4 m/s