Newton's Laws & Dynamics

Q1. (3 marks)

• Statement: A body remains at rest or in uniform straight-line motion unless acted on by a net external force. (1)
• Operational definition: force is that which changes a body's state of rest or uniform motion — i.e. that which produces acceleration. (1)
• The absence of acceleration (in an inertial frame) signals zero net force; force is defined/detected by the acceleration it causes. (1)

Q2. (4 marks)

• Impulse–momentum: $F\,t = \Delta p = m(v_f - v_i)$. (1)
• $\Delta p = 2(0 - 5) = -10\ \text{kg m/s}$. (1)
• $F = \Delta p / t = -10/0.4 = -25\ \text{N}$. (1)
• Magnitude $= 25\ \text{N}$ (opposite to motion). (1)

Q3. (5 marks) (a) FBD: weight $mg$ down, normal $N$ up, applied force $10\ \text{N}$ horizontal, friction $f$ opposing applied force. (2) (b) $N = mg = 4\times9.8 = 39.2\ \text{N}$. (1)

• Max static friction $= \mu_s N = 0.5\times39.2 = 19.6\ \text{N} > 10\ \text{N}$, so block stays still. (1)
• Actual friction $=$ applied force $= 10\ \text{N}$ (static friction adjusts to balance). (1)

Key point: friction is not automatically $\mu_s N$; it equals applied force up to the maximum.

Q4. (4 marks)

• On incline at angle $\theta$: forces along plane are $mg\sin\theta$ (down) and friction $f$ (up). Perpendicular: $N = mg\cos\theta$. (1)
• At angle of repose, block is on verge of sliding, so $f = f_{max} = \mu_s N$. (1)
• Equilibrium along plane: $mg\sin\theta = \mu_s N = \mu_s mg\cos\theta$. (1)
• Divide: $\tan\theta = \mu_s$. (1)

Q5. (5 marks)

• For $m_2$ (down): $m_2 g - T = m_2 a$; for $m_1$ (up): $T - m_1 g = m_1 a$. (1)
• Add: $(m_2 - m_1)g = (m_1 + m_2)a$. (1)
• $a = \dfrac{(m_2-m_1)g}{m_1+m_2} = \dfrac{(2)(9.8)}{8} = 2.45\ \text{m/s}^2$. (1)
• $T = m_1(g+a) = 3(9.8+2.45) = 36.75\ \text{N}$. (1)
• (Check: $T = m_2(g-a) = 5(7.35) = 36.75\ \text{N}$.) (1)

Q6. (4 marks) (a) Friction between tyres and road provides the centripetal force. (1) (b) $F_c = \dfrac{mv^2}{r} = \dfrac{1000\times15^2}{50}$ (1) $= \dfrac{225000}{50}$ (1) $= 4500\ \text{N}$. (1)

Q7. (5 marks)

• Energy conservation: KE at surface = magnitude of PE to reach infinity: $\tfrac12 m v_e^2 = \dfrac{GMm}{R}$. (1)
• Solve: $v_e = \sqrt{\dfrac{2GM}{R}}$. (1)
• Substitute: $v_e = \sqrt{\dfrac{2(6.67\times10^{-11})(6.0\times10^{24})}{6.4\times10^{6}}}$. (1)
• Numerator $= 8.004\times10^{14}$; divide by $6.4\times10^6 = 1.2506\times10^8$. (1)
• $v_e = \sqrt{1.25\times10^8} \approx 1.12\times10^{4}\ \text{m/s}$ ($\approx 11.2\ \text{km/s}$). (1)

Q8. (3 marks)

• Statement: For every action there is an equal and opposite reaction; forces occur in pairs acting on different bodies. (1)
• Misconception: that action–reaction pairs cancel out / produce equilibrium. (1)
• Why wrong: the two forces act on different bodies, so they never cancel on a single body; they cannot be added to give net force on one object. (1)

Q9. (4 marks)

• At the top, gravity + tension provide centripetal force: $mg + T = \dfrac{mv^2}{r}$. (1)
• Minimum speed when $T = 0$: $mg = \dfrac{mv_{min}^2}{r}$. (1)
• $v_{min} = \sqrt{gr}$. (1)
• $v_{min} = \sqrt{9.8\times0.8} = \sqrt{7.84} = 2.8\ \text{m/s}$. (1)

[
  {"claim":"Q2 force magnitude = 25 N","code":"m=2; v_i=5; v_f=0; t=Rational(4,10); F=m*(v_f-v_i)/t; result = (abs(F)==25)"},
  {"claim":"Q5 acceleration = 2.45 m/s^2 and tension = 36.75 N","code":"m1=3; m2=5; g=Rational(98,10); a=(m2-m1)*g/(m1+m2); T=m1*(g+a); result = (a==Rational(49,20)) and (T==Rational(147,4))"},
  {"claim":"Q6 centripetal force = 4500 N","code":"m=1000; v=15; r=50; Fc=m*v**2/r; result = (Fc==4500)"},
  {"claim":"Q9 minimum speed at top = 2.8 m/s","code":"g=Rational(98,10); r=Rational(8,10); v=sqrt(g*r); result = (simplify(v-Rational(28,10))==0)"},
  {"claim":"Q7 escape velocity approx 1.12e4 m/s","code":"G=6.67e-11; M=6.0e24; R=6.4e6; ve=sqrt(2*G*M/R); result = (abs(ve-1.12e4) < 200)"}
]