Physics Solution Q41

Question 41: Negative Mass Dynamics

Core Concept: Newton’s Laws with Negative Mass

The gravitational force is given by $F = \frac{G m_1 m_2}{r^2}$. The acceleration of a particle is $a = \frac{F}{m}$. Crucially, if mass $m$ is negative, the acceleration vector points in the direction opposite to the force vector.

Case (a): Both masses negative ($m_1 < 0, m_2 < 0$)

The product $m_1 m_2$ is positive, so the gravitational force is attractive (Force on 1 points to 2, Force on 2 points to 1). However, since masses are negative, acceleration is opposite to force.

$a_1$ points AWAY from 2.
$a_2$ points AWAY from 1.

Result: They move away from each other. (Matches q, r)

Case (b): $m_1 > 0, m_2 < 0$ and $|m_1| = |m_2|$

Product $m_1 m_2$ is negative. Force is repulsive.

Force on $m_1$ (pos) is away from $m_2$. Acc $a_1$ is away.
Force on $m_2$ (neg) is away from $m_1$. Acc $a_2$ is towards $m_1$ (opposite to force).

Particle 1 runs away, Particle 2 chases. Since masses are equal in magnitude, $|a_1| = |a_2|$. They move together with constant separation. Result: B follows A with constant separation. (Matches t)

Case (c): $m_1 > 0, m_2 < 0$ and $|m_1| > |m_2|$

Similar to case (b), 1 runs and 2 chases. However, since $|m_2|$ is smaller, $a_2 = F/|m_2|$ is larger than $a_1 = F/|m_1|$. The chaser (2) is faster than the runner (1). Result: B follows A and collides. (Matches s)

Case (d): $m_1 < 0, m_2 > 0$ and $|m_1| > |m_2|$

Force is repulsive.

Force on $m_1$ (neg) is away from $m_2$. Acc $a_1$ is towards $m_2$. ($m_1$ chases)
Force on $m_2$ (pos) is away from $m_1$. Acc $a_2$ is away from $m_1$. ($m_2$ runs)

Here 1 chases 2. Since $|m_1| > |m_2|$, the chaser’s acceleration $a_1$ is smaller than the runner’s acceleration $a_2$. The runner escapes. Result: Eventually B escapes away from A. (Matches r)

Correct Matching:
(a) → (q) and (r)
(b) → (t)
(c) → (s)
(d) → (r)