Newton's Laws & Dynamics

Question 1 — Atwood Machine

(a) FBD (2): For $m_1$: weight $m_1 g$ down, tension $T$ up. For $m_2$: weight $m_2 g$ down, tension $T$ up. (1 each)

(b) Derivation (5): Let $m_1$ accelerate down with $a$, $m_2$ up with $a$ (inextensible ⇒ same magnitude). $m_1 g - T = m_1 a \quad(1)$ $T - m_2 g = m_2 a \quad(1)$ Add: $(m_1 - m_2)g = (m_1 + m_2)a$ (1) $\boxed{a = \frac{(m_1 - m_2)g}{m_1 + m_2}}$ (1) Substitute back: $\boxed{T = \frac{2 m_1 m_2 g}{m_1 + m_2}}$ (1)

(c) Explanation (3):

• Tension is uniform because the string is massless and inextensible over a frictionless massless pulley — no net force can build across a massless element (1).
• $T < m_1 g$ because $m_1$ accelerates downward (net downward force needed) (1).
• $T > m_2 g$ because $m_2$ accelerates upward (net upward force needed) (1).

Question 2 — Angle of repose

(a) FBD & resolution (3): Weight $mg$ vertically down; Normal $N$ ⊥ surface; friction $f$ up the incline. Perpendicular: $N = mg\cos\theta$ (1.5) Along incline: $mg\sin\theta - f = ma$ (1.5)

(b) Derivation (4): At impending sliding, $a = 0$ and $f = f_{max} = \mu_s N$ (1). $mg\sin\theta_r = \mu_s N = \mu_s mg\cos\theta_r \quad(2)$ $\boxed{\tan\theta_r = \mu_s}\quad(1)$

(c) Numbers (3): $\mu_s = \tan 30° = \frac{1}{\sqrt3} \approx 0.577$ (2). The angle of repose equals the angle of friction (both = $\arctan\mu_s$) (1).

Question 3 — Banking of roads

(a) Frictionless derivation (6): FBD: weight $mg$ down, normal $N$ perpendicular to road (tilted by $\theta$). (1) Vertical: $N\cos\theta = mg$ (2) Horizontal (centripetal, toward center): $N\sin\theta = \dfrac{mv^2}{r}$ (2) Divide: $\boxed{\tan\theta = \frac{v^2}{rg}}\quad(1)$

(b) With friction, max speed (4): At $v_{max}$ friction acts down the slope (magnitude $\mu_s N$). Perpendicular: $N = mg\cos\theta + \dfrac{mv^2}{r}\sin\theta$ Along horizontal/centripetal balance gives: $\boxed{v_{max} = \sqrt{\frac{rg(\tan\theta + \mu_s)}{1 - \mu_s\tan\theta}}}\quad(4)$

(c) Numbers (2): $\tan 20° = 0.3640$. Numerator factor: $0.3640 + 0.4 = 0.7640$. Denominator: $1 - 0.4(0.3640) = 0.8544$. $v_{max} = \sqrt{80 \times 9.8 \times 0.7640/0.8544} = \sqrt{700.9} \approx 26.5\ \text{m/s}$. (2)

Question 4 — Vertical circular motion

(a) FBD (2): Top: tension $T_t$ down + weight $mg$ down. Bottom: tension $T_b$ up, weight $mg$ down. (1 each)

(b) Min speed at top (3): At top both point toward center: $T_t + mg = \frac{mv_{top}^2}{L}$ (1) Minimum: $T_t = 0$ (1): $\boxed{v_{top} = \sqrt{gL}}\quad(1)$

(c) Bottom speed (3): Energy conservation, height difference $2L$: $\tfrac12 m v_{bottom}^2 = \tfrac12 m v_{top}^2 + mg(2L)$ (1) $v_{bottom}^2 = gL + 4gL = 5gL$ (1) $\boxed{v_{bottom} = \sqrt{5gL}}\quad(1)$

(d) Explanation (2):

• Top: tension + gravity together provide centripetal force. (0.5)
• Bottom: tension minus gravity provides centripetal force. (0.5)
• Side: tension provides centripetal force; gravity is tangential. (1)

Question 5 — Escape & orbital velocity

(a) Escape velocity (4): Energy conservation, escape means reaching $r\to\infty$ with $v=0$ (total energy = 0): $\tfrac12 m v_{esc}^2 - \frac{GMm}{R} = 0$ (2) $\boxed{v_{esc} = \sqrt{\frac{2GM}{R}}}\quad(2)$

(b) Orbital velocity (3): Gravity provides centripetal force: $\frac{GMm}{r^2} = \frac{mv_{orb}^2}{r}$ (2) $\boxed{v_{orb} = \sqrt{\frac{GM}{r}}}\quad(1)$

(c) Relation (3): At surface $r = R$: $v_{esc} = \sqrt{2GM/R}$, $v_{orb} = \sqrt{GM/R}$ (1). $\frac{v_{esc}}{v_{orb}} = \sqrt{2} \Rightarrow v_{esc} = \sqrt{2}\,v_{orb}\quad(2)$

Question 6 — Pseudo-forces & weightlessness

(a) Elevator scale (4): FBD: normal $N$ up, weight $mg$ down. Upward acceleration $a$: $N - mg = ma$ (2) $\boxed{N = m(g + a)}$ (2) — apparent weight exceeds true weight.

(b) Explanation (4):

• In the freely-falling (orbiting) frame, a pseudo-force $-ma$ arises; since $a = g$ (free fall), it exactly cancels gravity, giving zero apparent weight (2).
• True weightlessness requires zero gravitational field (never actual near Earth); apparent weightlessness is what the astronaut feels — gravity acts fully but both astronaut and craft accelerate together so the normal/contact force vanishes (2).

[
  {"claim":"Atwood acceleration for m1=3,m2=1,g=9.8 equals 4.9","code":"m1,m2,g=3,1,9.8; a=(m1-m2)*g/(m1+m2); result = abs(a-4.9)<1e-9"},
  {"claim":"Atwood tension for m1=3,m2=1,g=9.8 equals 14.7","code":"m1,m2,g=3,1,9.8; T=2*m1*m2*g/(m1+m2); result = abs(T-14.7)<1e-9"},
  {"claim":"mu_s from angle of repose 30 deg is 1/sqrt(3)","code":"from sympy import tan,pi,sqrt,Rational,simplify; mu=tan(pi/6); result = simplify(mu-1/sqrt(3))==0"},
  {"claim":"Banked road vmax approx 26.5 m/s for r=80,theta=20,mu=0.4","code":"from sympy import tan,pi,sqrt; th=20*pi/180; mu=0.4; r=80; g=9.8; v=sqrt(r*g*(tan(th)+mu)/(1-mu*tan(th))); result = abs(float(v)-26.5)<0.2"},
  {"claim":"v_bottom = sqrt(5gL) from vtop=sqrt(gL) and height 2L","code":"from sympy import symbols,sqrt,simplify; g,L=symbols('g L',positive=True); vb2=g*L+4*g*L; result = simplify(sqrt(vb2)-sqrt(5*g*L))==0"},
  {"claim":"v_esc = sqrt(2)*v_orb at surface","code":"from sympy import symbols,sqrt,simplify; G,M,R=symbols('G M R',positive=True); vesc=sqrt(2*G*M/R); vorb=sqrt(G*M/R); result = simplify(vesc-sqrt(2)*vorb)==0"}
]