Question

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio 2 : 2 : 3. The two pieces of equal masses fly off perpendicular to each other with a speed of 18 m/s each. The velocity of the heavier fragment is ___ m/s.

• A. $12\sqrt{2}$
• B. 12
• C. $10\sqrt{2}$
• D. $24\sqrt{2}$

Hint:

Resultant of two perpendicular vectors $P$ is $P\sqrt{2}$.

Answer 1

The Correct Option is A

To solve this problem, we need to apply the principle of conservation of linear momentum. When the body explodes, the total initial momentum before the explosion is zero, as the body was initially at rest. After the explosion, the sum of the momenta of all fragments must also be zero to conserve momentum.

Let the masses of the three fragments be \(2x\), \(2x\), and \(3x\) respectively. Based on the given mass of the entire body (14 kg), we can express the mass of each fragment as follows:

\(2x + 2x + 3x = 14 \, \text{kg}\) 

Solving for \(x\):

\(7x = 14\) \(x = 2 \, \text{kg}\)

Thus, the masses of the fragments are: \(2x = 4 \, \text{kg}\), \(2x = 4 \, \text{kg}\), and \(3x = 6 \, \text{kg}\).

Let us determine the velocity of the heavier fragment (\(6 \, \text{kg}\)) after the explosion. Given that the two fragments of mass \(4 \, \text{kg}\) each fly off at right angles to each other with a speed of \(18 \, \text{m/s}\), let's consider their momentum vectors:

• Momentum of the first fragment = \(4 \times 18 = 72 \, \text{kg m/s}\) in the x-direction.
• Momentum of the second fragment = \(4 \times 18 = 72 \, \text{kg m/s}\) in the y-direction.

For momentum conservation, the third fragment’s momentum should balance these two:

\(\sqrt{(72)^2 + (72)^2} = \sqrt{2 \times 72^2} = 72\sqrt{2} \, \text{kg m/s}\)

Since momentum = mass × velocity, the velocity \(v\) of the third fragment is:

\(6v = 72\sqrt{2}\)

Solving for \(v\):

\(v = \frac{72\sqrt{2}}{6} = 12\sqrt{2} \, \text{m/s}\)

Thus, the velocity of the heavier fragment is \(12\sqrt{2} \, \text{m/s}\). Therefore, the correct answer is \(12\sqrt{2}\).