Level 1 — RecognitionNewton's Laws & Dynamics

Newton's Laws & Dynamics

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level 1: Recognition Test

Time Limit: 20 minutes Total Marks: 30 Instructions: Answer all questions. For True/False, a correct answer with no justification earns half marks.

Section A — Multiple Choice (1 mark each)

Q1. Newton's first law is essentially a statement about: (a) the equality of action and reaction (b) inertia and the operational definition of force (c) the conservation of energy (d) gravitational attraction

Q2. A body of mass $2\ \text{kg}$ experiences a net force of $10\ \text{N}$ . Its acceleration is: (a) $2\ \text{m/s}^2$ (b) $5\ \text{m/s}^2$ (c) $20\ \text{m/s}^2$ (d) $0.2\ \text{m/s}^2$

Q3. Which statement about an action–reaction pair is correct? (a) They act on the same body (b) They cancel each other out (c) They act on different bodies and never cancel (d) The action is always larger than the reaction

Q4. A block of mass $m$ rests on a horizontal table. The normal force on it equals: (a) always $mg$ (b) $mg$ only if no other vertical force acts (c) always zero (d) $2mg$

Q5. The maximum static friction force is given by: (a) $\mu_k N$ (b) $\mu_s N$ (c) $\mu_s mg\sin\theta$ (d) $\mu_k mg$

Q6. The angle of repose $\theta$ is related to the coefficient of static friction by: (a) $\mu_s = \sin\theta$ (b) $\mu_s = \cos\theta$ (c) $\mu_s = \tan\theta$ (d) $\mu_s = \cot\theta$

Q7. Escape velocity from Earth's surface (radius $R$ , mass $M$ ) is: (a) $\sqrt{GM/R}$ (b) $\sqrt{2GM/R}$ (c) $\sqrt{GM/2R}$ (d) $2\sqrt{GM/R}$

Q8. Gravitational potential energy of two masses separated by $r$ is: (a) $mgh$ (b) $+GMm/r$ (c) $-GMm/r$ (d) $-GMm/r^2$

Q9. In an ideal Atwood machine with masses $m_1 > m_2$ , the acceleration is: (a) $\dfrac{m_1+m_2}{m_1-m_2}g$ (b) $\dfrac{m_1-m_2}{m_1+m_2}g$ (c) $\dfrac{m_1}{m_2}g$ (d) $g$

Q10. The centripetal acceleration of an object in uniform circular motion (speed $v$ , radius $r$ ) is: (a) $v/r$ (b) $v^2/r$ directed toward the centre (c) $v^2/r$ directed outward (d) $vr$

Q11. An astronaut in a freely orbiting space station feels weightless because: (a) there is no gravity in orbit (b) the station and astronaut fall together with the same acceleration (c) the astronaut's mass is zero (d) air resistance cancels gravity

Section B — Matching (1 mark each pair, 6 marks)

Q12. Match each concept in Column X with its correct expression/description in Column Y.

Column X	Column Y
(i) Orbital velocity (circular orbit)	(P) $mv^2/r$ toward centre
(ii) Centripetal force	(Q) $\sqrt{GM/r}$
(iii) Banking angle (no friction)	(R) $-GMm/r$
(iv) Gravitational potential energy	(S) $\tan\theta = v^2/(rg)$
(v) Gravitational field intensity	(T) pseudo-force in a rotating frame
(vi) Centrifugal force	(U) $GM/r^2$

Section C — True/False with Justification (2 marks each: 1 for verdict, 1 for reason)

Q13. "The normal force on a block always equals its weight."

Q14. "In Newton's third law, since action equals reaction, the two forces cancel and no motion can occur."

Q15. "Kinetic friction is generally less than the maximum static friction between the same surfaces."

Q16. "The value of $g$ decreases as you go to higher altitude above Earth's surface."

Q17. "At the top of a vertical circle, the minimum speed for a ball on a string is $\sqrt{gr}$ ."

Q18. "Pseudo-forces are real forces that satisfy Newton's third law."

Q19. "A satellite in circular orbit has orbital speed independent of its own mass."

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) — Newton's first law defines inertia and gives an operational meaning to "force" as that which changes a body's state of motion. (1)

Q2. (b) $5\ \text{m/s}^2$ — $a = F/m = 10/2 = 5$ . (1)

Q3. (c) — Action and reaction act on different bodies, hence never cancel (cancellation requires same body). (1)

Q4. (b) — Normal force adjusts to balance vertical forces; equals $mg$ only when no other vertical force (e.g., no applied vertical push/pull) acts. (1)

Q5. (b) $\mu_s N$ — maximum static friction $= \mu_s N$ . (1)

Q6. (c) $\mu_s = \tan\theta$ — at repose, $mg\sin\theta = \mu_s mg\cos\theta \Rightarrow \mu_s=\tan\theta$ . (1)

Q7. (b) $\sqrt{2GM/R}$ — from $\tfrac12 mv^2 = GMm/R$ . (1)

Q8. (c) $-GMm/r$ — true gravitational PE with zero reference at infinity. (1)

Q9. (b) $\dfrac{m_1-m_2}{m_1+m_2}g$ — standard Atwood result. (1)

Q10. (b) $v^2/r$ toward the centre. (1)

Q11. (b) — Both fall (accelerate) together at the same rate, so no contact/normal force ⇒ apparent weightlessness. (1)

Section B (Q12, 1 mark per correct pair)

(i)→(Q); (ii)→(P); (iii)→(S); (iv)→(R); (v)→(U); (vi)→(T). (6)

Section C (2 marks each)

Q13. FALSE (1). Reason: Normal force equals $mg$ only on a horizontal surface with no other vertical forces; in a lift accelerating, on an incline, or with applied vertical forces it differs (e.g. $N = mg\cos\theta$ on incline). (1)

Q14. FALSE (1). Reason: The two forces act on different bodies, so they do not cancel on any single object; motion is governed by the net force on the body of interest. (1)

Q15. TRUE (1). Reason: Typically $\mu_k < \mu_s$ , so $\mu_k N < \mu_s N$ ; this is why it takes more force to start motion than to maintain it. (1)

Q16. TRUE (1). Reason: $g = GM/(R+h)^2$ decreases as altitude $h$ increases. (1)

Q17. TRUE (1). Reason: At minimum, tension $=0$ , so gravity provides centripetal force: $mg = mv^2/r \Rightarrow v=\sqrt{gr}$ . (1)

Q18. FALSE (1). Reason: Pseudo-forces exist only in non-inertial frames and have no reaction partner; they do not obey Newton's third law. (1)

Q19. TRUE (1). Reason: $v_{orb}=\sqrt{GM/r}$ depends only on central mass $M$ and radius $r$ , not the satellite's mass. (1)

[
  {"claim":"Q2 acceleration = 5 m/s^2","code":"F=10; m=2; result = (F/m == 5)"},
  {"claim":"Q9 Atwood acceleration formula sign/form","code":"m1,m2,g=symbols('m1 m2 g'); a=(m1-m2)/(m1+m2)*g; result = simplify(a - (m1-m2)*g/(m1+m2))==0"},
  {"claim":"Q7 escape velocity from KE=PE gives sqrt(2GM/R)","code":"G,M,R,m,v=symbols('G M R m v',positive=True); sol=solve(Eq(Rational(1,2)*m*v**2, G*M*m/R), v); result = sqrt(2*G*M/R) in sol"},
  {"claim":"Q17 min top-of-circle speed sqrt(g r)","code":"g,r,m,v=symbols('g r m v',positive=True); sol=solve(Eq(m*g, m*v**2/r), v); result = sqrt(g*r) in sol"}
]