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 m s$$^{-1}$$ and 40 m s$$^{-1}$$ respectively. The velocity of the third piece will be

A body of mass $$M$$ at rest explodes into three pieces with mass ratio 1 : 1 : 2. The two smaller pieces fly off perpendicular to each other with velocities 30 m s$$^{-1}$$ and 40 m s$$^{-1}$$.

Using conservation of momentum, let the masses be $$\frac{M}{4}$$, $$\frac{M}{4}$$, and $$\frac{M}{2}$$.

Initial momentum = 0, so final momentum must also be zero:

$$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$$

Calculating the momentum of the two smaller pieces gives

$$p_1 = \frac{M}{4} \times 30 = \frac{30M}{4}$$

$$p_2 = \frac{M}{4} \times 40 = \frac{40M}{4}$$

Since the two pieces move perpendicular to each other:

$$|\vec{p}_1 + \vec{p}_2| = \sqrt{p_1^2 + p_2^2} = \frac{M}{4}\sqrt{30^2 + 40^2} = \frac{M}{4}\sqrt{900 + 1600} = \frac{M}{4} \times 50$$

Therefore, the momentum of the third piece must satisfy

$$|\vec{p}_3| = |\vec{p}_1 + \vec{p}_2| = \frac{50M}{4}$$

Substituting into $$\frac{M}{2} \times v_3 = \frac{50M}{4}$$ and solving for $$v_3$$ yields

$$v_3 = \frac{50}{4} \times \frac{2}{1} = 25 \text{ m s}^{-1}$$

The correct answer is Option C: 25 m s$$^{-1}$$.