Question

A bomb of mass \(m\) at rest explodes into three parts. If these three parts move horizontally with equal speeds in different directions, then the masses of the three parts can be

• A. \(\frac{3m}{11},\frac{m}{3},\frac{13m}{33}\)
• B. \(\frac{m}{6},\frac{m}{3},\frac{m}{2}\)
• C. \(\frac{4m}{19},\frac{5m}{19},\frac{10m}{19}\)
• D. \(\frac{6m}{29},\frac{8m}{29},\frac{15m}{29}\)

Hint:

For explosion problems always conserve momentum. If initial momentum is zero, vector sum of all final momenta must also be zero.

Answer 1

The Correct Option is B

Concept: Initial momentum is zero because bomb is at rest. Hence after explosion total momentum must remain zero. For three vectors with equal magnitudes of velocity to give zero resultant, masses must satisfy triangle law. So masses should be capable of forming sides of triangle.

Step 1: Check option B Masses \[ \frac m6,\frac m3,\frac m2 \] Multiply by common factor 6 \[ 1,2,3 \] Triangle condition \[ 1+2=3 \] Possible limiting equilibrium. Thus acceptable. Other options fail triangle condition. Hence answer \[ \boxed{\frac m6,\frac m3,\frac m2} \]