JEE Mains · Physics · STD 11 - 3.2 motion in plane
An object is projected with kinetic energy K from a point A at an angle \(60^{\circ}\) with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :
- A 1 : 2
- B 2 : 3
- C 1 : 4
- D 3 : 4
Answer & Solution
Correct Answer
(D) 3 : 4
Step-by-step Solution
Detailed explanation
\( (KE)_{A}=K=\frac{1}{2}mu^{2} \) \(( KE )_{ B }=\frac{ K }{4}=\frac{1}{2} m\left(\frac{ u }{2}\right)^2=\frac{ K }{4}\left( u _{ B }= u \cos 60^{\circ}=\frac{ u }{2}\right)\) \( (KE)_{C}=K \) Ratio \(= \frac{K-K/4}{K} \) \(=\frac{3 K / 4}{K}=\frac{3}{4}\)
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