A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms⁻¹ and 40 ms⁻¹ respectively. The velocity of the third piece will be :

Solution

Physics sir · 4 steps

The body starts at rest, so the third piece must cancel the vector sum of the two small momenta.

1. Zero total momentum. Take the masses as m, m and 2m. The body was at rest, so the total momentum after the explosion is still zero. The third piece must carry momentum equal and opposite to the combined momentum of the two small pieces.
2. Add perpendicular momenta. The small pieces have momenta 30m and 40m at right angles to each other. Add them as vectors, not as plain numbers.

   p = √((30m)² + (40m)²) = 50m

3. Third piece. The third piece has mass 2m, so 2m × v = 50m, which gives v = 25 m s⁻¹. It moves opposite to that resultant.
4. So. The third piece moves at 25 m s⁻¹, option (b).

The idea to keep. In an explosion from rest, the momentum vectors of all the pieces add to zero. Add perpendicular momenta by Pythagoras, then divide by the mass of the remaining piece.

Answer from the NTA final answer key. Solution by PrepWiser.