A particle slides down a frictionless parabolic (y + x2 ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) Height at P = height at A
(c) Total energy at P = total energy at A
(d) Time of travel from A to B = time of travel from B to P.
This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- c
Explanation– as the given track y=x2 is a frictionless track thus total energy will be same throughout the journey.
Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P
Hence (KE)B = (KE)P
Total energy at A = PE= total energy at B = KE= total energy at P
= PE+KE
Potential energy at A is converted to KE and PE at P hence
(PE)P< (PE)A
Hence (height)P= (height)A
As height of p < height of A
Hence path length AB > path length BP
<p>This is a Multiple Choice Questions as classified in NCERT Exemplar</p><p><strong>Answer- c</strong></p><p><strong>Explanation</strong>– as the given track y=x<sup>2</sup> is a frictionless track thus total energy will be same throughout the journey.</p><p>Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P</p><p>Hence (KE)<sub>B </sub>= (KE)<sub>P</sub></p><p>Total energy at A = PE= total energy at B = KE= total energy at P</p><p>= PE+KE</p><p>Potential energy at A is converted to KE and PE at P hence</p><p> (PE)P< (PE)A</p><p>Hence (height)P= (height)A</p><p>As height of p < height of A</p><p>Hence path length AB > path length BP</p>
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