Physics Galaxy Discussion Questions: Solutions

Question

Solution

Discussion Question 1

Two particles A and B start moving from the same point on a straight line. A moves with constant speed (v_0). B starts from rest with constant acceleration (a). They meet twice. Find the condition for the second meeting time and the ratio of their speeds at the first meeting.

Solution (Reasoning Approach)

Let (t = 0) be the start.
Position of A: (x_A = v_0 t)
Position of B: (x_B = \frac12 a t^2)

They meet when (x_A = x_B)
[ v_0 t = \frac12 a t^2 \quad\Rightarrow\quad t\left(v_0 - \fraca2 t\right) = 0 ]
Solutions: (t = 0) (initial point) and (t = \frac2v_0a) (first meeting).

But they meet twice? That means B must overtake A, then A overtakes B again — impossible for these equations unless direction changes.
So the hidden trick: One of them changes direction after some time (e.g., B accelerates, decelerates, or A reverses).
Correct interpretation: If B has constant acceleration and A constant speed, they meet only once after start. So “meet twice” implies either B first goes backward or motion is on a circle. physics galaxy discussion questions solutions

Thus the discussion reveals: This question is actually from circular motion — two runners on a circular track starting together.

Let’s fix: Track length (L), A speed (v_0), B from rest with acc (a) along track.
Positions (angle):
[ \theta_A = \fracv_0 tR, \quad \theta_B = \frac12 \fracaR t^2 ]
Meeting means (\theta_A - \theta_B = 2n\pi) or (v_0 t - \frac12 a t^2 = n L) (with (L = 2\pi R)).

Two meetings after (t=0) ⇒ two positive roots of ( \frac12 a t^2 - v_0 t + n L = 0 ) for (n=1) and maybe (n=2) depending on parameters.

Condition for exactly two meetings (excluding start): Discriminant > 0, and smallest root for n=1 the first meeting, second root for n=1 same as first root for n=0? No — that’s degenerate. Better: The physics galaxy discussion would lead to:
[ t_1 = \fracv_0 - \sqrtv_0^2 - 2aLa, \quad t_2 = \fracv_0 + \sqrtv_0^2 - 2aLa \quad\text(for n=1) ]
And second meeting (n=2) possible if ( \sqrtv_0^2 - 4aL ) real ⇒ condition.

Moving from mechanics to electrostatics, the questions become even more philosophical.

Question

Solution

The Physics Galaxy series provides a solution set (often in a separate booklet or online portal). Use it like a chess master uses a solver:

If you don't know the answer, push the variables to infinity or zero.

Question:
Why do elliptical galaxies have much larger $M/L$ ratios than spiral disks? What does this imply about their formation?

Solution:

Discussion Question 4

An ideal gas undergoes a cycle consisting of:
1→2: Isochoric heating,
2→3: Isothermal expansion,
3→1: Isobaric compression.
Given (P_1, V_1, T_1) and (P_2 = 2P_1, V_2 = V_1), find efficiency of the cycle.

Solution

Let’s set (P_1 = P_0, V_1 = V_0, T_1 = T_0).
1→2: (V) const, (P) doubles ⇒ (T_2 = 2T_0).
Heat added (Q_12 = n C_v (T_2 - T_1) = n C_v (2T_0 - T_0) = n C_v T_0).

2→3: Isothermal at (T=2T_0). (P_3) from 3→1: isobaric ⇒ (P_3 = P_1 = P_0).
At point 2: (P_2 = 2P_0, V_2 = V_0) ⇒ (nR(2T_0) = 2P_0 V_0) ⇒ (nRT_0 = P_0 V_0).
At point 3: (P_3 = P_0, T_3 = 2T_0) ⇒ (V_3 = \fracnR(2T_0)P_0 = \frac2P_0 V_0P_0 = 2V_0).

Work in 2→3: (W_23 = nR(2T_0) \ln\fracV_3V_2 = 2nRT_0 \ln 2).
Heat in 2→3: (Q_23 = W_23) (isothermal) = (2 P_0 V_0 \ln 2).

3→1: Isobaric compression at (P_0) from (V_3=2V_0) to (V_1=V_0).
Work (W_31 = P_0 (V_1 - V_3) = P_0(V_0 - 2V_0) = -P_0 V_0).
Heat rejected (Q_31 = n C_p (T_1 - T_3) = n C_p (T_0 - 2T_0) = -n C_p T_0).

Efficiency (\eta = \fracW_netQ_in = \fracQ_12+Q_23+Q_31Q_12+Q_23).
But (Q_31) negative, so
(W_net = Q_12 + Q_23 + Q_31 = nC_v T_0 + 2P_0 V_0 \ln 2 - n C_p T_0)
But (n C_p T_0 = n (C_v + R) T_0 = n C_v T_0 + n R T_0 = n C_v T_0 + P_0 V_0).

So (W_net = n C_v T_0 + 2P_0 V_0 \ln 2 - n C_v T_0 - P_0 V_0)
(W_net = P_0 V_0 (2\ln 2 - 1)).

(Q_in = Q_12 + Q_23 = n C_v T_0 + 2P_0 V_0 \ln 2).
But (n C_v T_0 = \fracP_0 V_0\gamma - 1). Question

Thus
[ \eta = \frac2\ln 2 - 1\frac1\gamma-1 + 2\ln 2 ]
For monatomic gas (\gamma=5/3) ⇒ (\frac1\gamma-1 = 1.5) ⇒ (\eta \approx \frac0.38631.5 + 1.3863 \approx 0.134) (13.4%).

Discussion: Compare with Carnot between (T_0) and (2T_0): (\eta_Carnot = 1 - T_1/T_2 = 0.5). Large difference due to irreversible isochoric heating and isobaric cooling.