Question

A body of mass 1 kg initially at rest explodes into three fragments of masses in the ratio 1:1:3. The two pieces of equal masses fly off perpendicular to each other with a speed of 15 $ms^{-1}$
What is the velocity of the heavier fragment?
$\begin{aligned} & A.\text{ }10\sqrt{2}m{{s}^{-1}} \\\ & B.\text{ }5\sqrt{3}m{{s}^{-1}} \\\ & C.\text{ }10\sqrt{3}m{{s}^{-1}} \\\ & D.\text{ }5\sqrt{2}m{{s}^{-1}} \\\ \end{aligned}$

Answer 1

Apply the conservation of the momentum. The principal says that the total initial momentum vectors are the same as the total final momentum vector. So first find out the initial momentum and then find the final vector momentum with the assumed values and calculate the unknowns then.

Complete step by step answer:
The mass of the body M initially is M is 1 kg
Then the mass of the first fragment is $m_1 = \dfrac{1}{5} \times 1 = 0.2 kg$
The mass of the second fragment is $m_2 = \dfrac{1}{5} \times 1 = 0.2 kg$
The mass of the third fragment is ${{m}_{3}}$ =$\dfrac{3}{5} \times 1 = 0.6 kg$
We are given the velocity of the two fragments ${{m}_{1}}\text{ and }{{m}_{2}}$ is v = 15 $m/s$
So the initial momentum becomes zero as the initial value of the velocity is zero
So now let's calculate the final momentum of the all the fragments
We get the value of the final fragments as below:
The momentum is defined as the product of mass time’s velocity and hence we get the value of the momentum of the first particle as:
${{p}_{1}}={{m}_{1}}v=0.2\times 15=3kgm/s$
The momentum of the second particle as
${{p}_{2}}={{m}_{2}}v=0.2\times 15=3kgm/s$
Now the third particle flows in a direction perpendicular to the momentum of the first and second particle so we get the value of the momentum of the third particle as
$p_3 = \sqrt{p_1^2 + p_2^2}$
The value of the momentum of the third particle is obtained and hence the value of the speed of the third particle can be found as: ${{p}_{3}}=\sqrt{{{3}^{2}}+{{3}^{2}}}=3\sqrt{2}kgm/{{s}^{-1}}$ .
Thus the velocity of the third piece =${{v}_{3}}=\dfrac{{{p}_{3}}}{{{m}_{3}}}=\dfrac{3\sqrt{2}}{0.6}=5\sqrt{2}m{{s}^{-1}}$
Hence the correct answer is option (d).

Note:
One of the main possibilities of making mistakes is finding the value of the momentum of the third particle. We need to consider the vector direction of the momentum of the other particle and find the vector direction of the third particle to find its momentum.