Question

A body of mass \[5{\text{ }}kg\] explodes at rest into three fragments with masses in the ratio \[1:1:3\]. The fragments with equal masses fly in mutually perpendicular directions with speeds of \[21{\text{ }}m/s\]. The velocity of heaviest fragment in \[m/s\] will be
A. $7\sqrt 2 $
B. $5\sqrt 2 $
C. $3\sqrt 2 $
D. $\sqrt 2 $

Answer 1

Hint: In this question, we are given that a body is explodes into three fragments at rest where the mass \[5{\text{ }}kg\] distributed in the ratio of \[1:1:3\]. Also, the fragment of equal mass flies in perpendicular direction with the speed \[21{\text{ }}m/s\]. Here, the resultant of the momentum of the same mass fragment will be equal to the momentum of the third fragment (conservation of the momentum).

Formula used:
Formula of momentum –
$P = mv$ , where $m,v$ are the mass and velocity of the object respectively.
Magnitude of resultant momentum –
${P_{net}} = \sqrt {{{\left( {{m_1}{v_1}} \right)}^2} + {{\left( {{m_2}{v_2}} \right)}^2}} $

Complete step by step solution:
Given that the body explodes with the mass \[5{\text{ }}kg\] into three fragments in the ratio \[1:1:3\]. Therefore, the mass of each fragment will be $1{\text{ }}kg,1{\text{ }}kg,3{\text{ }}kg$ respectively.
It implies that, ${m_1} = 1{\text{ }}kg,{m_2} = 1{\text{ }}kg,{m_3} = 3{\text{ }}kg$

Now, the fragment of same mass flies with the speed \[21{\text{ }}m/s\]
\[ \Rightarrow {v_1} = 21{\text{ }}m/s,{v_2} = 21{\text{ }}m/s\]
Applying the formula of momentum $P = mv$,
Momentum of first (fragment) $ = {m_1}{v_1}$
$\Rightarrow 1 \times 21$
$\Rightarrow 21{\text{ }}kg{\text{ }}m/\sec $
Momentum of second (fragment) $ = {m_2}{v_2}$
$\Rightarrow 1 \times 21$
$\Rightarrow 21{\text{ }}kg{\text{ }}m/\sec $

Now, resultant momentum when first and second fragment are flying mutually perpendicular direction,
${P_{net}} = \sqrt {{{\left( {{m_1}{v_1}} \right)}^2} + {{\left( {{m_2}{v_2}} \right)}^2}} $
$\Rightarrow {P_{net}} = \sqrt {{{\left( {21} \right)}^2} + {{\left( {21} \right)}^2}} $
${P_{net}} = 21\sqrt 2 {\text{ }}kg{\text{ }}m/\sec $
According to the conservation law of linear momentum, the momentum of the third fragment is equal to the resultant momentum.
$ \Rightarrow {m_3}{v_3} = {P_{net}} = 21\sqrt 2 {\text{ }}kg{\text{ }}m/\sec $
\[ \Rightarrow 3 \times {v_3} = 21\sqrt 2 \]
$\therefore {v_3} = 7\sqrt 2 {\text{ }}kg{\text{ }}m/\sec $

Hence, option A is the correct answer.

Note: To solve such questions, one should always remember the linear momentum’s conservation law. Also, the formula of net momentum when the bodies of same mass fly mutually perpendicular to each other is ${P_{net}} = \sqrt {{{\left( {{m_1}{v_1}} \right)}^2} + {{\left( {{m_2}{v_2}} \right)}^2} + 2\left( {{m_1}{v_1}} \right)\left( {{m_2}{v_2}} \right)\cos \theta } $ when the value of $\theta $ is ${90^ \circ }$. As we know that, $\cos {90^ \circ } = 0$. Therefore, we used formula as ${P_{net}} = \sqrt {{{\left( {{m_1}{v_1}} \right)}^2} + {{\left( {{m_2}{v_2}} \right)}^2}} $. If angle will be other than ${90^ \circ }$ then, use ${P_{net}} = \sqrt {{{\left( {{m_1}{v_1}} \right)}^2} + {{\left( {{m_2}{v_2}} \right)}^2} + 2\left( {{m_1}{v_1}} \right)\left( {{m_2}{v_2}} \right)\cos \theta } $ and put the value of angle.