Question

A body of mass 1 kg, initially at rest, explodes and breaks into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other with a speed of 30 m/sec each. If the velocity of the heavier fragment is $10\sqrt{x}$, determine x.

Answer 1

Hint: According to Newton's second law for a particular system rate of change of momentum will be equal to the external force acting on the system. The rock has exploded into 3 fragments and this will be due to internal force hence at the instant of explosion we can consider no external impulsive force acting and solve this problem

Formula used: $p = mv$
If ${F_{ext}} = 0$ then ${p_i} = {p_f}$

Complete step by step solution:
It is given that the mass of the body is 1 kg and exploded in the ratio 1:1:3. Let us consider the initial mass to be ${m_i}$ and after the explosion it got divided into three masses ${m_1},{m_2},{m_3}$.
Since three masses are in 1:1:3 ratio two of them will be equal. Let ${m_1},{m_2}$ are equal.
So ${m_1} = \dfrac{1}{{1 + 1 + 3}} = \dfrac{1}{5}kg,{m_2} = \dfrac{1}{{1 + 1 + 3}} = \dfrac{1}{5}kg,{m_3} = \dfrac{3}{{1 + 1 + 3}} = \dfrac{3}{5}kg$
Now let us conserve the momentum of the body
Momentum is the product of mass and velocity
Its initial momentum ${p_i}$ = 0 as body is initially at rest
It is given in the question that two equal masses moved mutually perpendicular directions with speed of 30m/s. let ${v_1}$ be the velocity of first mass and ${v_2}$ be the velocity of the second mass and ${v_3}$ be the velocity of the third mass.
After the explosion let the momentum of the first mass be ${p_1}$ and momentum of the second mass be ${p_2}$ and momentum of the third mass be ${p_3}$. Let the first mass had moved in $\mathop n\limits^ \wedge $ direction while the second mass moved in $\mathop o\limits^ \wedge $ direction. $\mathop n\limits^ \wedge $ and $\mathop o\limits^ \wedge $ are mutually perpendicular.
Now final momentum is ${p_f} = {p_1} + {p_2} + {p_3}$
${p_1} = \dfrac{1}{5} \times 30\mathop n\limits^ \wedge = 6\mathop n\limits^ \wedge $
${p_2} = \dfrac{1}{5} \times 30\mathop o\limits^ \wedge = 6\mathop o\limits^ \wedge $
${p_3} = \dfrac{3}{5} \times {v_3}$
By doing vector addition of ${p_1},{p_2}$ we get ${p_1} + {p_2} = \sqrt {{6^2} + {6^2}} = 6\sqrt 2 $
We have ${p_i} = {p_f}$ i.e
 $\eqalign{
  & 0 = {p_1} + {p_2} + {p_3} \cr
  & \left| {{p_3}} \right| = \left| {{p_1} + {p_2}} \right| \cr
  & \left| {\dfrac{3}{5} \times {v_3}} \right| = 6\sqrt 2 \cr
  & \left| {{v_3}} \right| = 10\sqrt 2 \cr} $

Hence, the value of x = 2.

Note: The following things should be on finger tips in order to solve the question:
(1) Definition of momentum along with resultant.
(2) Formula of mass in case of ratio: $\dfrac{\text{ratio of broken pieces}}{\text{total ratio}}\times \text{Total mass before collision}$.
(3) Formula of momentum: ${{P}_{\text{resultant}}}=\sqrt{{{P}_{1}}+{{P}_{2}}}$.
(4) In the solution, the resultant is not taken as negative even the direction of momentum of G is opposite to it. This is because the net momentum must be zero and this can be done by the fact that the two resultants either are equal to zero or they give a zero when subtracted.