Question

A stationary body of mass $3kg$ explodes into three equal pieces. Two of the pieces fly off at right angles to each other. One with a velocity of $2{\text{i}}\,{\text{m}}{{\text{s}}^{ - 1}}$ and the other with a velocity of $3{\text{j}}\,{\text{m}}{{\text{s}}^{ - 1}}$. If the explosion takes place in ${10^{ - 5}}{\text{s}}$, the average force acting on the third piece in newtons is:
A. $\left( {2{\text{i + 3j}}} \right) \times {10^{ - 5}}$
B. $- \left( {2{\text{i + 3j}}} \right) \times {10^5}$
C. $\left( {{\text{3i + 2j}}} \right) \times {10^5}$
D. $- \left( {2{\text{i + 3j}}} \right) \times {10^{ - 5}}$

Answer 1

To solve this problem first the mass of each piece is equal and moving with some velocity. Now conserving the momentum we can find the momentum of the third piece. Then after using the force momentum relation formula where time is also given to use we can find the average force acting on the third piece.

Complete step by step answer:
As per the problem there is a stationary body of mass $3kg$ explodes into three equal pieces. Two of the pieces fly off at right angles to each other. One with a velocity of $2{\text{i}}\,{\text{m}}{{\text{s}}^{ - 1}}$ and the other with a velocity of $3{\text{j}}\,{\text{m}}{{\text{s}}^{ - 1}}$. The explosion takes place in ${10^{ - 5}}{\text{s}}$.We need to calculate the average force acting on the third piece in newtons.

As the body explodes into three equal pieces there mass must be equal.
Mass of the body before it explodes, $M = 3\,kg$.
After collision the mass of the three pieces will be,
$m = m_1 = m_2 = m_3 = \dfrac{M}{3}$
$\Rightarrow m = m_1 = m_2 = m_3 = \dfrac{{3Kg}}{3}$
Hence we will get,
$m = m_1 = m_2 = m_3 = 1kg$

Let the momentum of the third piece be $\overrightarrow {P_3}$ and that of the other two be $\overrightarrow {P_1}$ and $\overrightarrow {P_2}$.Now we can write momentum as a product of mass and velocity.Where velocity of one piece is $\overrightarrow {v1} = 2{\text{i}}\,{\text{m}}{{\text{s}}^{ - 1}}$ and the velocity of other piece is $\overrightarrow {v2} = 3{\text{j}}\,{\text{m}}{{\text{s}}^{ - 1}}$.
Now converting the momentum we will get,
$\overrightarrow {Pi} = \overrightarrow {Pf}$
Where the initial momentum is zero.

Now we can write the above equation as,
$0 = \overrightarrow {Pf}$
Final moment is the sum of all the three momum of the exploded piece.
$0 = \overrightarrow {P_1} + \overrightarrow {P_2} + \overrightarrow {P_3} \ldots \ldots \left( 1 \right)$
Now,
$\overrightarrow {P_1} = m\overrightarrow {v1}$
Putting the know value we will get,
$\overrightarrow {P_1} = 1kg \times 2{\text{i}}\,m{s^{ - 1}}$
$\Rightarrow \overrightarrow {P_1} = 2{\text{i}}\,kg\,m{s^{ - 1}}$

Now,
$\overrightarrow {P_2} = m\overrightarrow {v2}$
Putting the know value we will get,
$\overrightarrow {P_2} = 1kg \times 3{\text{j}}\,m{s^{ - 1}}$
$\Rightarrow \overrightarrow {P_2} = 3{\text{j}}\,kg\,m{s^{ - 1}}$
Now putting the values in the equation $\left( 1 \right)$ we will get,
$0 = 2{\text{i}}\,{\text{kgm}}{{\text{s}}^{ - 1}} + 3{\text{j}}\,{\text{kgm}}{{\text{s}}^{ - 1}} + \overrightarrow {P_3}$
Taking the constant value to other side we will get,
$- \left( {2{\text{i}}\,{\text{kgm}}{{\text{s}}^{ - 1}} + 3{\text{j}}\,{\text{kgm}}{{\text{s}}^{ - 1}}} \right) = \overrightarrow {P_3} \ldots \ldots \left( 2 \right)$

Now applying average force formula on the third piece we will get,
$\overrightarrow F = \dfrac{{\overrightarrow {P_3} }}{t}$
We know,
$t = {10^{ - 5}}{\text{s}}$
Now putting equation $\left( 2 \right)$ and time in the average force formula we will get,
$\overrightarrow F = \dfrac{{ - \left( {2{\text{i}}\,{\text{kgm}}{{\text{s}}^{ - 1}} + 3{\text{j}}\,{\text{kgm}}{{\text{s}}^{ - 1}}} \right)}}{{{{10}^{ - 5}}}}$
Taking the denominator value to the numerator we will get,
$\overrightarrow F = - \left( {2{\text{i}}\,{\text{kgm}}{{\text{s}}^{ - 1}} + 3{\text{j}}\,{\text{kgm}}{{\text{s}}^{ - 1}}} \right) \times {10^5}$
$\therefore \overrightarrow F = - \left( {2{\text{i}} + 3{\text{j}}} \right) \times {10^5}\,N$

Therefore the correct option is $\left( B \right)$.

Note: Initial momentum of the body is zero because initial velocity before the explosion is zero as the body is at rest. In the second equation there is no need to write the momentum of the third piece as a product of mass and velocity as the average force is calculated by dividing momentum with time and also it will make the solution a bit lengthy.