Question

A bomb exploited at rest in three fragments of equal mass two fragments fly of right angle to each other with velocities 9m/s and 12m/s calculate the velocity of third fragments

Answer 1

-21 m/s perpendicular to first piece. this is to maintain the conservation of momentum.

Answer 2

Answer:

Velocity of the third fragment is - 15√2 m/s.

Explanation:

Given;-

       Masses of particles, = m ( same )

      Velocity of Particle₁ , v₁= 9 m/s  

      Velocity of Particle₂, v₂= 12 m/s

Let the velocity of the 3rd particle be v m/s.

Now,

Since the bomb initially was at rest, its net initial momentum is zero. And from the concept of conservation of momentum, the net final momentum will also be zero. So,

                              Pi = Pf = 0

              P₁ + P₂ + P₃ = Pf = 0

        m(9i) + m(12 j) + m(v) = 0

(Refer to the attached image)

          m( 9i + 12j + v) = 0

                   15√2 + v = 0                

                   v = - 15√2 m/s

Hope it helps! ;-))