Newton's Laws of Motion
ICSE · Class 11 · Physics
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Quick Quiz: Newton's Laws of Motion
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A force of 20 N acts on a body of mass 4 kg initially at rest on a smooth horizontal surface. What is the acceleration produced in the body?
A cricket ball of mass 0.15 kg moving at 20 m/s is brought to rest by a batsman in 0.05 s. What is the magnitude of the average force exerted by the batsman on the ball?
Two masses m₁ = 3 kg and m₂ = 5 kg are connected by a string and placed on a smooth horizontal surface. A force F = 16 N is applied on m₂. What is the tension in the string connecting them?
In an Atwood machine, two masses m₁ = 6 kg and m₂ = 4 kg are connected by a string over a smooth pulley. What is the acceleration of the system? (Take g = 10 m/s²)
Sample Questions
A person of mass 60 kg is standing in a lift accelerating upward at 3 m/s². What is his apparent weight? (Take g = 10 m/s²)
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780 N
Step 1: Identify the situation — lift accelerates upward, so apparent weight > real weight. Step 2: Formula: W_apparent = m(g + a) when lift accelerates upward. Step 3: W_apparent = 60 × (10 + 3) = 60 × 13 = 780 N. Why wrong options are incorrect: - 420 N: Used m(g - a) = 60 × 7, formula for downward acceleration. - 600 N: This is actual weight mg = 60 × 10, ignores acceleration. - 180 N: Only calculated m × a = 60 × 3, forgot to add gravity.
A gun of mass 5 kg fires a bullet of mass 50 g with a velocity of 200 m/s. What is the recoil velocity of the gun?
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2 m/s
Step 1: Convert mass of bullet: m_bullet = 50 g = 0.05 kg. Mass of gun M = 5 kg. Step 2: Apply conservation of linear momentum. Initial momentum = 0 (both at rest). Step 3: Final momentum: m_bullet × v_bullet + M × V_gun = 0. 0.05 × 200 + 5 × V_gun = 0. Step 4: 10 + 5 × V_gun = 0 → V_gun = -10/5 = -2 m/s. Magnitude of recoil velocity = 2 m/s. Why wrong options are incorrect: - 1 m/s: Forgot to convert grams to kg. - 4 m/s: Divided bullet velocity by gun mass directly (4 = 200/50). - 0.5 m/s: Calculated 0.05/5 × 200 = 0.02 × 200 incorrectly.
Which of the following is the correct statement of Newton's First Law of Motion?
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A body at rest remains at rest and a body in uniform motion continues in uniform motion unless acted upon by an external unbalanced force.
Newton's First Law (Law of Inertia) states that an object at rest stays at rest and an object in uniform motion stays in uniform motion along a straight line unless compelled by an external unbalanced force. Why wrong options are incorrect: - Option A is incomplete — it ignores the case of a body at rest and also doesn't specify 'unbalanced' force. - Option C is Newton's Third Law (Law of Action-Reaction). - Option D is Newton's Second Law (F = dp/dt).
A block of mass 10 kg is placed on a smooth inclined plane making an angle of 30° with the horizontal. What is the acceleration of the block down the plane? (Take g = 10 m/s², sin 30° = 0.5)
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5 m/s²
Step 1: On a smooth inclined plane, only the component of gravity along the plane causes acceleration. Step 2: The component along the plane = mg sin θ. Step 3: Applying Newton's second law: mg sin θ = ma → a = g sin θ. Step 4: a = 10 × sin 30° = 10 × 0.5 = 5 m/s². Why wrong options are incorrect: - 10 m/s²: This is 'g' itself, ignoring the incline angle. - 8.66 m/s²: This uses g cos 30° = 10 × 0.866, which is the component perpendicular to the plane (normal reaction direction). - 2.5 m/s²: This is g sin 30° / 2, an incorrect halving of the answer.
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