Answer

Verified

52.8k+ views

**Hint:**In this solution, we will use the law of conservation of linear momentum. In the absence of an external force, the net momentum of a system remains conserved in all directions.

Formula used: In this solution, we will use the following formula:

Momentum of an object $\vec p = m\vec v$ where $m$ is the mass and $\vec v$ is the velocity of the object.

**Complete step by step answer:**

When the shell at rest at origin explodes into three fragments, there is no external force acting on the shell. This implies that the net momentum of the system should remain zero, that is the net momentum of all the three fragments should add up to zero.

Now the two fragments with mass 1 and 2 kg respectively fly off with speeds \[12m/s\] along the x-axis and \[8m/s\] along the y-axis respectively.

To conserve momentum, the momentum of the first two masses should be equal to the momentum of the third mass.

Let us do this by calculating the net momentum of the masses of 1 and 2 kg respectively.

The momentum of those two masses will be

$\overrightarrow {{P_{1,2}}} = (1)(12)\hat i + (2)(8)\hat j$

$ \Rightarrow \overrightarrow {{P_{1,2}}} = 12\hat i + 16\hat j$

The magnitude of this momentum will be

$\left| {\overrightarrow {{P_{1,2}}} } \right| = \sqrt {{{12}^2} + {{16}^2}} = 20$

This net momentum should be equal to the momentum of the third mass to conserve the net momentum of the system i.e. $\left| {\overrightarrow {{P_{1,2}}} } \right| = \left| {\overrightarrow {{P_3}} } \right|$. Since the third mass moves with a velocity of \[40m/s\], its mass can be calculated as

\[{m_3} = \dfrac{{\left| {\overrightarrow {{P_3}} } \right|}}{{\left| {{v_3}} \right|}}\]

\[ \Rightarrow {m_3} = \dfrac{{20}}{{40}} = 0.5\,kg\]

Hence the mass of the third fragment is $0.5\,kg$. So the mass of the entire shell will be

\[M = 1 + 2 + 0.5\]

**\[ \Rightarrow M = 3.5\,kg\] which corresponds to option (C).**

**Note:**While calculating the momentum of the third fragment using the momentum of the first and the second fragment, we should directly take the magnitude of their net momentum as we’ve directly been given the velocity of the third fragment and we’re not concerned with its direction but only the magnitude of its momentum.

Recently Updated Pages

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Which of the following amino acids is an essential class 12 chemistry JEE_Main

Which of the following is least basic A B C D class 12 chemistry JEE_Main

Out of the following hybrid orbitals the one which class 12 chemistry JEE_Main

Other Pages

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Vant Hoff factor when benzoic acid is dissolved in class 12 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main