Question

A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off at right angles to each other , one with velocity 2i^m/s2i^m/s and the other with velocity 3j^m/s3j^m/s. If the explosion takes place in 10−5s,10−5s, the average force acting on the third piece in Newtons is

(a)(2i^+3j^)×10−5(c)(3i^−2j^)×105(b)−(2i^+3j^)×10+5(d)(2i^−3j^)×

​

Answer 1

Force= rate of change of momentum by momentum conservation

1(2i^)+1(3j^)+1(v¯)=01(2i^)+1(3j^)+1(v¯)=0

Δp=−[2i^+3j^]Δp=−[2i^+3j^]

F=ΔpΔtF=ΔpΔt

=−[2i^+3j^]×105N


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Answer 2

answer : option (b) -(2i + 3j) × 10^5 N

using law of conservation of linear momentum,

external force acting on the body is zero.

so, initial linear momentum = final linear momentum

⇒3kg × 0 = 1kg × 2i m/s + 1kg × 3j m/s + 1kg × v

⇒v = -(2i + 3j) m/s

now, momentum of third piece is P = -(2i + 3j) Kgm/s

we know, force = momentum/ time taken

so, force acting on the third fragment is = -(2i + 3j) / 10^-5 N = -(2i + 3j) × 10^5 N

hence, answer is -(2i + 3j) × 10^5 N

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