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Question: An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angle...

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m/s and the second part of mass 2 kg moves with 8 m/s speed. If the third part flies off with 4 m/s speed, then its mass is
(A) 3 kg
(B) 5 kg
(C) 7 kg
(D) 17 kg

Explanation

Solution

Here we are given a rock initially at rest and then suddenly an explosion occurs due to which it breaks into three parts in a horizontal plane. Explosion did not occur due to any external forces but it happened due to internal forces. Since, no external force acts on the system, momentum of the system remains conserved.

Complete step by step answer:
Initially the rock was at rest and we do not know the total mass of the rock. Initial momentum of the system is zero because momentum is the product of mass and velocity and here initial velocity is zero.
Initial momentum of the system= 0 kgm/s
After the explosion the rock breaks into three parts. The masses of the rocks are 1kg, 2 kg and m kg respectively. Their velocities are 12 m/s, 8 m/s and 4 m/s respectively. First two go off at right angles to each other.
Final momentum of the system is p1+p2+p3\overrightarrow{{{p}_{1}}}+\overrightarrow{{{p}_{2}}}+\overrightarrow{{{p}_{3}}}
Applying the law of conservation of momentum, we get,
(1×12)i^+(2×8)j^+p3=0(1\times 12)\widehat{i}+(2\times 8)\widehat{j}+\overrightarrow{{{p}_{3}}}=0

& \overrightarrow{{{p}_{3}}}=-(12\widehat{i}+16\widehat{j}) \\\ & =\sqrt{{{12}^{2}}+{{16}^{2}}} \\\ & =20 \\\ \end{aligned}$$ But 4m=20 So, m= 5kg **So, the correct option is (B)** **Note:** If an external force is acting on the system, we would not be able to solve this problem using conservation of linear momentum law. Momentum is a vector quantity because it is the product of mass and velocity and velocity is a vector quantity.