NEET Laws of Motion — Set 12

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Question 1

A $ 1800 \, \text{kg} $ truck accelerates at $ 2.5 \, \text{m/s}^2 $ for $ 4 \, \text{s} $, then a $ 2 \, \text{kg} $ stone is dropped. What is the horizontal distance traveled by the stone in $ 1.5 \, \text{s} $? (Neglect air resistance)

TopperSquare Expert · Accepted Answer

Truck’s velocity: $ v = u + at = 0 + 2.5 \times 4 = 10 \, \text{m/s} $. Stone inherits this horizontal velocity. No horizontal acceleration after drop. Distance: $ s = v \times t = 10 \times 1.5 = 15 \, \text{m} $.

Question 2

A $ 3.5 \, \text{kg} $ block on a $ 30^\circ $ incline ($ \mu_k = 0.25 $) is pulled upward by a $ 6.5 \, \text{kg} $ mass over a pulley. What is the tension? (Take $ g = 10 \, \text{m/s}^2 $, $ \sin 30^\circ = 0.5 $, $ \cos 30^\circ = 0.866 $)

TopperSquare Expert · Accepted Answer

For $ 6.5 \, \text{kg} $: $ 6.5g - T = 6.5a \Rightarrow 65 - T = 6.5a $. For $ 3.5 \, \text{kg} $: $ T - mg \sin 30^\circ - f_k = 3.5a $. $ N = mg \cos 30^\circ = 3.5 \times 10 \times 0.866 = 30.31 \, \text{N} $. Friction: $ f_k = 0.25 \times 30.31 \approx 7.58 \, \text{N} $. $ mg \sin 30^\circ = 35 \times 0.5 = 17.5 \, \text{N} $. Net force: $ T - 17.5 - 7.58 = 3.5a \Rightarrow T - 25.08 = 3.5a $. Solve: $ 65 - T = 6.5a $, $ T - 25.08 = 3.5a $. Substitute: $ 65 - (3.5a + 25.08) = 6.5a \Rightarrow 65 - 25.08 - 3.5a = 6.5a \Rightarrow 39.92 = 10a $. $ a \approx 3.99 \, \text{m/s}^2 $, $ T - 25.08 = 3.5 \times 3.99 \Rightarrow T - 25.08 \approx 13.97 \Rightarrow T \approx 39 \, \text{N} $.

Question 3

A $ 6 \, \text{kg} $ block on a $ 30^\circ $ incline ($ \mu_k = 0.25 $) is connected to a $ 3 \, \text{kg} $ mass over a pulley. What is the acceleration of the system? (Take $ g = 10 \, \text{m/s}^2 $, $ \sin 30^\circ = 0.5 $, $ \cos 30^\circ = 0.866 $)

TopperSquare Expert · Accepted Answer

Assume $ 6 \, \text{kg} $ moves down the incline, pulling $ 3 \, \text{kg} $ up. For $ 6 \, \text{kg} $: $ mg \sin 30^\circ - f_k - T = 6a $. $ N = mg \cos 30^\circ = 6 \times 10 \times 0.866 = 51.96 \, \text{N} $. Friction: $ f_k = 0.25 \times 51.96 \approx 12.99 \, \text{N} $. $ mg \sin 30^\circ = 60 \times 0.5 = 30 \, \text{N} $. Net force: $ 30 - 12.99 - T = 6a \Rightarrow 17.01 - T = 6a $. For $ 3 \, \text{kg} $: $ T - 3g = 3a \Rightarrow T - 30 = 3a $. Solve: $ 17.01 - T = 6a $, $ T - 30 = 3a $. Substitute: $ 17.01 - (3a + 30) = 6a \Rightarrow 17.01 - 30 - 3a = 6a \Rightarrow -12.99 = 9a $. $ a \approx -1.44 \, \text{m/s}^2 $ (opposite, $ 3 \, \text{kg} $ descends, $ a \approx 1.44 \, \text{m/s}^2 $).

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NEET Physics: Laws of Motion — MCQ Test 12

Q6. A \( 1000 \, \text{kg} \) car on a banked road (\( \theta = 15^\circ \), \( \mu_s = 0.3 \)) turns at radius \( 40 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Q7. A \( 6.8 \, \text{kg} \) block on a \( 30^\circ \) incline (\( \mu_k = 0.15 \)) is pulled upward by a \( 8.8 \, \text{kg} \) mass over a pulley. What is the acceleration? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.866 \))

Q8. A \( 0.2 \, \text{kg} \) ball moving at \( 30 \, \text{m/s} \) at \( 45^\circ \) to a wall rebounds at the same speed and angle. What is the impulse perpendicular to the wall? (Take \( \sin 45^\circ = \cos 45^\circ = 0.707 \))

Q9. A \( 2500 \, \text{kg} \) rocket accelerates upward at \( 3 \, \text{m/s}^2 \) with a thrust of \( 35000 \, \text{N} \). What is the mass ejection rate if the gas speed is \( 100 \, \text{m/s} \) relative to the rocket? (Take \( g = 10 \, \text{m/s}^2 \))