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JEECUP (UP Polytechnic) Laws of Motion — practice questions — page 3

Explore 75 free Laws of Motion questions in JEECUP (UP Polytechnic) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Laws of Motion questions and exercises with answers

Showing questions 61–75 of 75.

A lift descends with a downward acceleration equal to g. The apparent weight of a passenger standing inside is:Read the answer and worked solution →A 0.2 kg ball strikes the floor vertically at 10 m s⁻¹ and rebounds at 8 m s⁻¹. The magnitude of the impulse imparted by the floor on the ball is:Read the answer and worked solution →The optimum (design) speed for a car on a frictionless road banked at 45° along a curve of radius 100 m is (g = 10 m s⁻²):Read the answer and worked solution →True or false: what Aristotle got wrongRead the answer and worked solution →Complete Newton's first lawRead the answer and worked solution →Which law is this?Read the answer and worked solution →MomentumRead the answer and worked solution →True or false: inertia and forceRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Tap every statement the first law licensesRead the answer and worked solution →Flip and pair: name with contributionRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Order the historical argumentRead the answer and worked solution →

Worked practice questions

A lift descends with a downward acceleration equal to g. The apparent weight of a passenger standing inside is:

  1. zero
  2. twice mg
  3. equal to mg
  4. half of mg
Show answer and solution

Correct answer: A. zero

Apparent weight N = m(g − a). With downward acceleration a = g, N = 0 — the passenger is in free fall and feels weightless.

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A 0.2 kg ball strikes the floor vertically at 10 m s⁻¹ and rebounds at 8 m s⁻¹. The magnitude of the impulse imparted by the floor on the ball is:

  1. 3.6 N s
  2. 2.0 N s
  3. 1.6 N s
  4. 0.4 N s
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Correct answer: A. 3.6 N s

Taking upward as positive, impulse = m(v_up − v_down) = 0.2[8 − (−10)] = 0.2×18 = 3.6 N s.

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The optimum (design) speed for a car on a frictionless road banked at 45° along a curve of radius 100 m is (g = 10 m s⁻²):

  1. 31.6 m s⁻¹
  2. 10.0 m s⁻¹
  3. 45.0 m s⁻¹
  4. 22.4 m s⁻¹
Show answer and solution

Correct answer: A. 31.6 m s⁻¹

On a frictionless banked curve the design speed satisfies tanθ = v²/(rg), so v = √(rg tanθ) = √(100×10×tan45°) = √1000 ≈ 31.6 m s⁻¹.

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