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A particle slides down a frictionless parabolic $\left(y=x^2\right)$ track $(A-B-C)$ starting from rest at point $A$ (figure). 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
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The correct answers are:
total energy at $P=$ total energy at $A$
total energy at $P=$ total energy at $A$
According to the question, the given track $y=x^2$ is a frictionless track thus, total energy will be same throughout the journey.
By the law of conservation of energy.
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 $\mathrm{PE}$.
So, $(\mathrm{KE})_B>(\mathrm{KE})_P$
Total energy at $A=\mathrm{PE}$
$=$ Total energy at $B=\mathrm{KE}$
$=$ Total energy at $P$
$=\mathrm{PE}+\mathrm{KE}$
The potential energy at $A$, is converted to $\mathrm{KE}$ and $\mathrm{PE}$ at $P$.
Then, (PE) $P < (\mathrm{PE}) A$
So, (Height) $P < $ (Height) $A$
and as, height of $P < $ height of $A$.
So, path length $A B>$ path length $B P$
Thus, time of travel from $A$ to $B$
$\neq$ Time of travel from $B$ to $P$.
By the law of conservation of energy.
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 $\mathrm{PE}$.
So, $(\mathrm{KE})_B>(\mathrm{KE})_P$
Total energy at $A=\mathrm{PE}$
$=$ Total energy at $B=\mathrm{KE}$
$=$ Total energy at $P$
$=\mathrm{PE}+\mathrm{KE}$
The potential energy at $A$, is converted to $\mathrm{KE}$ and $\mathrm{PE}$ at $P$.
Then, (PE) $P < (\mathrm{PE}) A$
So, (Height) $P < $ (Height) $A$
and as, height of $P < $ height of $A$.
So, path length $A B>$ path length $B P$
Thus, time of travel from $A$ to $B$
$\neq$ Time of travel from $B$ to $P$.
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