Newton's Laws of Motion — Homi Bhabha Class 9 Physics MCQs with Solutions
Free Homi Bhabha Class 9 Physics Newton's Laws of Motion MCQs with step-by-step solutions (10 questions). Part of Laws of Motion & Force. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Newton's Laws of Motion · medium · numerical
A force acts on a body of mass 5 kg producing an acceleration of 3 m/s². The magnitude of the force is:
A. 1.67 N
B. 8 N
C. 15 N ✓ Correct
D. 45 N
Solution: By Newton's second law $F = ma = 5 \times 3 = 15$ N.
Q2 — Newton's Laws of Motion · medium · theory
When a moving bus stops suddenly, a standing passenger lurches forward. This is best explained by:
A. Newton's first law (inertia of motion) ✓ Correct
B. Newton's third law
C. the law of conservation of momentum
D. Newton's second law
Solution: The lower body stops with the bus while the upper body tends to continue moving forward due to inertia of motion.
Q3 — Newton's Laws of Motion · hard · theory
Newton's second law of motion states that the net force on a body equals:
A. the rate of change of its momentum ✓ Correct
B. the product of its mass and velocity
C. its change in momentum
D. its mass times its speed
Solution: $F = \dfrac{\Delta p}{\Delta t}$; for constant mass this reduces to $F = ma$.
Q4 — Newton's Laws of Motion · hard · numerical
A force of 10 N acts to the right and a force of 4 N acts to the left on a 2 kg body. Its acceleration is:
A. 7 m/s² to the right
B. 3 m/s² to the right ✓ Correct
C. 5 m/s² to the right
D. 2 m/s² to the left
Solution: Net force $= 10 - 4 = 6$ N; $a = \dfrac{F}{m} = \dfrac{6}{2} = 3$ m/s² in the direction of the larger force.
Q5 — Newton's Laws of Motion · medium · theory
Assertion (A): A body in uniform motion continues to move uniformly unless an external force acts on it. Reason (R): This is a statement of Newton's first law (law of inertia).
A. Both A and R are true but R is NOT the correct explanation of A
B. A is false but R is true
C. A is true but R is false
D. Both A and R are true and R is the correct explanation of A ✓ Correct
Solution: A is exactly the content of the first law, which R names — so R correctly explains A.
Q6 — Newton's Laws of Motion · medium · numerical
A body of mass 4 kg experiences a net force of 20 N. Its acceleration is:
A. 80 m/s²
B. 5 m/s² ✓ Correct
C. 0.2 m/s²
D. 16 m/s²
Solution: $a = \dfrac{F}{m} = \dfrac{20}{4} = 5$ m/s².
Q7 — Newton's Laws of Motion · easy · theory
Newton's first law of motion is also commonly known as the:
A. law of momentum
B. law of action and reaction
C. law of acceleration
D. law of inertia ✓ Correct
Solution: The first law states that a body stays at rest or in uniform motion unless acted on by a net external force, which is exactly the idea of inertia, so it is called the law of inertia.
Q8 — Newton's Laws of Motion · medium · numerical
A net force of 12 N acts on a body of mass 3 kg. The acceleration produced is:
A. 0.25 m/s²
B. 4 m/s² ✓ Correct
C. 9 m/s²
D. 36 m/s²
Solution: By Newton’s second law a = F/m = 12/3 = 4 m/s².
Q9 — Newton's Laws of Motion · hard · theory
A horse pulls a cart and both accelerate forward, even though the cart pulls back on the horse with an equal and opposite force. This is possible because:
A. the action and reaction cancel out, so no force is really needed to move
B. the horse’s pull on the cart is larger than the cart’s pull on the horse
C. the action and reaction act on different bodies, and the forward push of the ground on the horse’s hooves exceeds the backward resistance on the cart ✓ Correct
D. the reaction force acts a moment later than the action force
Solution: Action and reaction act on different bodies and never cancel on the same body. The horse pushes the ground back and the ground pushes it forward; when this exceeds the resistance on the cart, the whole system accelerates.
Q10 — Newton's Laws of Motion · hard · numerical
A body of mass 2 kg moving at 6 m/s is brought to rest in 3 s by a constant force. The magnitude of the force is:
A. 4 N ✓ Correct
B. 3 N
C. 2 N
D. 12 N
Solution: Acceleration a = (0 - 6)/3 = -2 m/s²; force magnitude F = m|a| = 2 × 2 = 4 N.