Newton's Laws & Dynamics

Question 1 (10)

(a) Driving force = weight of hanging mass $= m_2 g = 3.0\times9.8 = 29.4\ \text{N}$. Maximum static friction $= \mu_s m_1 g = 0.40\times4.0\times9.8 = 15.68\ \text{N}$. (1) Since $29.4 > 15.68$, the driving force exceeds max static friction → system moves. (2)

(b) With motion, kinetic friction $f_k = \mu_k m_1 g = 0.25\times4.0\times9.8 = 9.8\ \text{N}$. (1) Equations:

• $m_2$: $m_2 g - T = m_2 a$
• $m_1$: $T - f_k = m_1 a$ (1)

Add: $m_2 g - f_k = (m_1+m_2)a$ $a = \frac{m_2 g - \mu_k m_1 g}{m_1+m_2} = \frac{29.4-9.8}{7.0} = \frac{19.6}{7.0} = 2.8\ \text{m/s}^2$ (2) Tension: $T = m_2(g-a) = 3.0(9.8-2.8) = 3.0\times7.0 = 21.0\ \text{N}$. (1)

(c) Because $m_2$ is accelerating downward ($a\neq0$), net force on it must be non-zero; hence $T < m_2 g$. Tension equals weight only in equilibrium (static/constant velocity). (2)

Question 2 (12)

(a) FBD: weight $mg$ down, normal $N\perp$ surface, friction $f$ along surface (down the slope at max speed, since car tends to slide up). (1) Resolve (vertical & horizontal, centripetal = horizontal inward):

• Vertical: $N\cos\theta = mg + f\sin\theta$
• Horizontal: $N\sin\theta + f\cos\theta = \dfrac{mv^2}{R}$ (2)

At maximum speed $f = \mu_s N$. Substitute: (1) $N\cos\theta - \mu_s N\sin\theta = mg \Rightarrow N=\frac{mg}{\cos\theta-\mu_s\sin\theta}$ $N(\sin\theta+\mu_s\cos\theta)=\frac{mv^2}{R}$ Divide: (1) $\boxed{v_{\max}=\sqrt{Rg\,\frac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}}=\sqrt{Rg\,\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}}}$ (1)

(b) $\theta=20^\circ$: $\tan20^\circ=0.3640$. Numerator $=0.3640+0.30=0.6640$; Denominator $=1-0.30\times0.3640=0.8908$. (1) Ratio $=0.7454$. $Rg=80\times9.8=784$. (1) $v_{\max}=\sqrt{784\times0.7454}=\sqrt{584.4}=24.2\ \text{m/s}$. (1)

(c) No-friction speed: $v_0=\sqrt{Rg\tan\theta}=\sqrt{784\times0.3640}=\sqrt{285.4}=16.9\ \text{m/s}$. (1) Below $v_0$: car tends to slide down, friction acts up the slope. (1) Above $v_0$: car tends to slide up, friction acts down the slope. (1)

Question 3 (10)

(a) On inside of a track, only gravity + normal (inward) supply centripetal force. Minimum when $N=0$: $mg=\frac{mv_{\text{top}}^2}{r}\Rightarrow v_{\text{top}}=\sqrt{gr}=\sqrt{9.8\times0.5}=2.21\ \text{m/s}$ (3)

(b) A threaded bead has the wire on both sides; the wire can supply an outward (upward at top) reaction. So gravity need not fully provide centripetal force — the wire makes up any deficit. Thus $v_{\text{top}}$ can be arbitrarily small (minimum speed $\to 0$). (2)

(c) Released from rest at height of centre (level with axis), drop to bottom = distance $r$ below centre. (1) Energy conservation: $\tfrac12 m v^2 = mg r$ (2) $v=\sqrt{2gr}=\sqrt{2\times9.8\times0.5}=\sqrt{9.8}=3.13\ \text{m/s}$ (2)

Question 4 (10)

(a) Gravity provides centripetal force: $\frac{GMm}{r^2}=\frac{mv_o^2}{r}\Rightarrow v_o=\sqrt{\frac{GM}{r}}$ (3)

(b) $r=R_p+h=6.4\times10^6+3.6\times10^5=6.76\times10^6\ \text{m}$. (1) $GM=6.67\times10^{-11}\times6.0\times10^{24}=4.002\times10^{14}$. (1) $v_o=\sqrt{4.002\times10^{14}/6.76\times10^6}=\sqrt{5.92\times10^{7}}=7.69\times10^{3}\ \text{m/s}\approx7.7\ \text{km/s}$. (1)

(c) Escape: total energy zero: $\tfrac12 mv_{\text{esc}}^2=\frac{GMm}{R_p}\Rightarrow v_{\text{esc}}=\sqrt{\frac{2GM}{R_p}}$. (1) $=\sqrt{2\times4.002\times10^{14}/6.4\times10^6}=\sqrt{1.251\times10^{8}}=1.118\times10^{4}\ \text{m/s}\approx11.2\ \text{km/s}$. (1) Surface orbital speed $v_s=\sqrt{GM/R_p}$, so $v_{\text{esc}}=\sqrt{2GM/R_p}=\sqrt2\sqrt{GM/R_p}=\sqrt2\,v_s$. (2)

Question 5 (8)

(a) In the carriage frame the bob is in equilibrium under: tension $T$ along string, weight $mg$ down, pseudo-force $ma$ backward (opposite to acceleration). (1) Horizontal: $T\sin\phi = ma$; Vertical: $T\cos\phi = mg$. (2) Divide: $\tan\phi = a/g$. (1)

(b) Forward acceleration is horizontal only; vertical equilibrium unchanged, so normal force $N=mg$ → scale reads the same, $60g$. (Horizontal acceleration doesn't affect vertical normal force.) (2)

(c) $\tan\phi = a/g = 2.0/9.8 = 0.204 \Rightarrow \phi=11.5^\circ$. (Bob swings toward front of train during braking.) (2)

[
  {"claim":"Q1 acceleration = 2.8 m/s^2","code":"a=(3.0*9.8-0.25*4.0*9.8)/7.0; result=abs(a-2.8)<0.05"},
  {"claim":"Q1 tension = 21.0 N","code":"a=(3.0*9.8-0.25*4.0*9.8)/7.0; T=3.0*(9.8-a); result=abs(T-21.0)<0.1"},
  {"claim":"Q2 vmax approx 24.2 m/s","code":"th=pi/180*20; v=sqrt(80*9.8*(tan(th)+0.30)/(1-0.30*tan(th))); result=abs(float(v)-24.2)<0.3"},
  {"claim":"Q4 escape = sqrt(2)*surface orbital speed","code":"GM=6.67e-11*6.0e24; Rp=6.4e6; vesc=sqrt(2*GM/Rp); vs=sqrt(GM/Rp); result=abs(float(vesc)-sqrt(2)*float(vs))<1"},
  {"claim":"Q3 bottom speed = 3.13 m/s","code":"v=sqrt(2*9.8*0.5); result=abs(float(v)-3.13)<0.02"}
]