Topic: System of Masses
Concepts: Conservation of energy, Conservation of linear momentum

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Solution:

1st statement: If a particle at rest explodes into 2 particles, by conserving momentum and energy we obtain

The first equation tells us the magnitudes of the momenta are equal $p_1=p_2 \equiv p$ so

which can be solved for $p$ and then we know $K_1$ and $K_2$. In this case, we have two equations for two unknowns $p_1, p_2$. For three particles, we still have these two equations for three unknowns $p_1, p_2, p_3$ so there is not a unique solution. Hence, $N_1=2$.

2nd statement: If a particle at rest explodes into 2 particles, then by conservation of momentum,

so the velocities of the particles lie in a line (special case of lying in a plane). If we added some momentum $\vec{p}_0$ (perpendicular to the line) to the first particle, then we could restore momentum conservation by adding a third particle with momentum $-\vec{p}_0$. Thus, this counterexample illustrates the velocities of 3 particles do not have to lie in a line. Hence, $N_2=2$.

3rd statement: If a particle at rest explodes into 3 particles, then by conservation of momentum,

The plane containing $\vec{p}_2$ and $\vec{p}_3$ can be defined by the perpendicular vector $\vec{n}=\vec{p}_2 \times \vec{p}_3$. Using the fact that $\vec{n} \cdot \vec{p}_2=\vec{n} \cdot \vec{p}_3=0$, we have

so $\vec{p}_1$ is also perpendicular to $\vec{n}$ and thus in the plane of $\vec{p}_2$ and $\vec{p}_3$. Hence, the velocities of the particles lie in a plane. If we added some momentum $\vec{p}_0$ (perpendicular to the plane) to the first particle, then we could restore momentum conservation by adding a fourth particle with momentum $-\vec{p}_0$. Thus, this counterexample illustrates the velocities of 4 particles do not have to lie in a plane. Hence $N_3=3$.

Combining our results, the answer is C.