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Intermediate — requires understanding of vector dot products, conservation laws, and ability to apply formulas in collision problems.
A force of 20 N acts on a body at an angle of 60° to displacement of 4 m. The work done is: A. 40 J B. 80 J C. 160 J D. 20 J Answer: A Explanation: ( W = Fs\cos\theta = 20 \times 4 \times \cos60^\circ = 80 \times 0.5 = 40 \text{ J} ). Why others fail: B ignores the cosine factor and uses ( W = Fs ) directly.
A 2 kg block moves with speed 5 m/s. Its kinetic energy is: A. 10 J B. 25 J C. 50 J D. 100 J Answer: B Explanation: ( K = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 25 = 25 \text{ J} ). Why others fail: C results from forgetting the ( \frac{1}{2} ) in kinetic energy formula.
In a perfectly inelastic collision between two objects of equal mass moving toward each other with equal speed, the final kinetic energy is: A. Equal to initial B. Half of initial C. Zero D. One-fourth of initial Answer: C Explanation: Objects stick together and come to rest due to equal and opposite momenta; total momentum is zero, so final velocity is zero. Why others fail: B is chosen by those who confuse with energy loss in general inelastic collisions.
A ball falls from height ( h ) and rebounds to height ( 0.64h ). The coefficient of restitution is: A. 0.64 B. 0.8 C. 0.9 D. 0.7 Answer: B Explanation: ( e = \sqrt{\frac{h'}{h}} = \sqrt{0.64} = 0.8 ). Why others fail: A is chosen by those who use height ratio directly instead of square root.
A 10 kg mass moving at 6 m/s collides elastically with a stationary 2 kg mass. The final velocity of the 2 kg mass is: A. 8 m/s B. 10 m/s C. 12 m/s D. 14 m/s Answer: B Explanation: Using ( v_2 = \frac{2m_1u_1}{m_1 + m_2} = \frac{2 \times 10 \times 6}{10 + 2} = \frac{120}{12} = 10 \text{ m/s} ). Why others fail: A results from incorrect application of momentum conservation without considering elastic collision formula.
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