Question

A shell at rest at origin explodes into three fragments of masses 1kg, 2kg and m kg. The 1kg and 2kg fly off with speed of 12m/s along x-axis and 16m/s along y-axis respectively. If the m kg piece flies off with speed of 40m/s, the total mass of the shell must be
A) 3.8 kg
B) 4 kg
C) 4.5 kg
D) 5kg

Answer 1

In the question it is given that the shell explodes at rest at origin. Hence from the principle of conservation of momentum, we can imply that the net change in the momentum i.e. initial momentum is equal to final momentum which is equal to zero. Therefore using the above principle, we can accordingly determine the mass m which will enable us to determine the total mass of the shell.

Formula used:
${{P}_{initial}}={{P}_{final}}$
$P=mv$

Complete step-by-step solution:
Let us say we have a particle of mass ‘m’ and moves with velocity v. Therefore the momentum (P) of the particle is given by,
$P=mv$
Let us say a particle has some initial momentum ${{P}_{initial}}$. Further let us say if the particle encounters an obstacle such that it does not lose energy, then the final momentum ${{P}_{final}}$from the principle of conservation of momentum is given by,
${{P}_{initial}}={{P}_{final}}$
In the above question we are given that the shell at rest at origin explodes into three fragments of masses 1kg, 2kg and m kg. The 1kg and 2kg fly off with speed of 12m/s along x-axis and 16m/s along y-axis respectively. Hence the momentum ${{P}_{x}}$ of the particle along the x-axis and along the y-axis ${{P}_{y}}$ is given by,
$\begin{aligned} & {{P}_{x}}=m{{v}_{x}} \\\ & \because m=1kg,\text{ }{{v}_{x}}=12m{{s}^{-1}} \\\ & \Rightarrow {{P}_{x}}=1kg\times 12m{{s}^{-1}} \\\ & \therefore {{P}_{x}}=12kgm{{s}^{-1}} \\\ \end{aligned}$
Similarly the momentum of the particle along the y-axis is,
$\begin{aligned} & {{P}_{y}}=m{{v}_{y}} \\\ & \because m=2kg,\text{ }{{v}_{y}}=16m{{s}^{-1}} \\\ & \Rightarrow {{P}_{y}}=2kg\times 16m{{s}^{-1}} \\\ & \therefore {{P}_{y}}=32kgm{{s}^{-1}} \\\ \end{aligned}$

Since momentum is a vector quantity, from the above diagram, the net momentum (P) of the particle along x and y is,
$\begin{aligned} & \overline{P}=\overline{{{P}_{x}}}+\overline{{{P}_{y}}} \\\ & \Rightarrow \left| P \right|=\sqrt{{{\left( {{P}_{x}} \right)}^{2}}+{{\left( {{P}_{y}} \right)}^{2}}} \\\ & \Rightarrow \left| P \right|=\sqrt{{{\left( 12 \right)}^{2}}+{{\left( 32 \right)}^{2}}} \\\ & \therefore \left| P \right|=34.2kgm{{s}^{-1}} \\\ \end{aligned}$
In the question it is given that the shell explodes at rest. Hence the above momentum should be equal to momentum of particles of mass ‘m’ moving with velocity ${{v}_{m}}=40m{{s}^{-1}}$. Hence we obtain,
$\begin{aligned} & P=m{{v}_{m}} \\\ & \Rightarrow 34.2kgm{{s}^{-1}}=m40m{{s}^{-1}} \\\ & \therefore m=\dfrac{34.2kgm{{s}^{-1}}}{40m{{s}^{-1}}}=0.855kg \\\ \end{aligned}$
Hence the total mass M of the above shell is equal to,
$\begin{aligned} & M=1kg+2kg+0.855kg \\\ & \therefore M=3.85kg\approx 3.8kg \\\ \end{aligned}$
Therefore the correct answer of the above question is option A.

Note: It is to be noted that when the shell explodes at rest, it implies that the final momentum is equal to zero. This is because classically we are not concerned of the momentum of the individual constituents inside the shell. We just have to consider the initial and final momentum, without considering the cause of the explosion.