A stationery body of mass 3kg explodes into three equal pieces. Two of the pieces fly off at right angles to each Other. One with a velocity of 2i^ m/s and other with a velocity of 3i^ m/s.If the explosion takes place in 10-5 s, the average force acting on the third piece in newtons is

Since the body explodes into three equal parts, therefore m1=m2=m3= m3=1 kgLet the velocity of the third part be  v→.According to the principle of conservation of linear momentum,Momentum of system before explosion = Momentum of system after explosionOr mv= m1v1+m2v2+m3v3Or  3×0=1×2i^+1x3j^+1×v→Or v=  - ( 2i^+3j^)m/sAverage force acting on the third particle is F→=mv→t=−1×(2i^+3j^)10−5=−(2i^+3j^)×105N