Question: 20: A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2: 2: 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is
(1) 2 \sqrt{2} v
(2) 3 \sqrt{2} v
(3) v
(4) \sqrt{2} v
Answer: Option (1)
Explanation:
Since the shell is initially at rest, the total momentum of the system before explosion is zero.
Let the masses of the three fragments be in the ratio 2:2:1.
The total mass is 5 parts, so the actual masses are
m_1 = \frac{2m}{5}, \quad m_2 = \frac{2m}{5}, \quad m_3 = \frac{m}{5}.
The two equal-mass fragments m_1 and m_2
move along mutually perpendicular directions with speed v.
Their momenta are p_1 = \frac{2m}{5}v and p_2 = \frac{2m}{5}v.
Since these momenta are perpendicular, the resultant momentum of these two fragments is
p = \sqrt{p_1^2 + p_2^2} = \sqrt{\left(\frac{2m}{5}v\right)^2 + \left(\frac{2m}{5}v\right)^2}.
p = \frac{2m}{5}v\sqrt{2}.
To conserve momentum, the momentum of the third fragment must be equal in magnitude and opposite in direction to this resultant momentum.
Thus, for the third fragment,
\frac{m}{5}u = \frac{2m}{5}v\sqrt{2},
where u is the speed of the lighter fragment.
Solving, u = 2\sqrt{2}v.
Hence, the speed of the third fragment is 2\sqrt{2}v.