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# A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg, and m kg. The fragments of masses 1 kg and 2 kg fly off with speeds $12m/s$along the x-axis and $8m/s$along the y-axis. If the m kg piece flies off with a speed of $40m/s$then find the total mass of the shell.A) 4 kgB) 5 kgC) $3.5\,kg$D) $4.5\,kg$

Last updated date: 17th Jun 2024
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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.

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.