Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A small roller coaster starts at point A with a speed u on a curved track as shown in the figure.
seo images

The friction between the rollercoaster and the track is negligible and it always remains in contact with the track. The speed of roller coaster at point D on the track will be:
A. $\sqrt {{u^2} + gh} $
B. $\sqrt {{u^2} + 2gh} $
C. $\sqrt {{u^2} + 4gh} $
D. $u$

seo-qna
Last updated date: 09th Sep 2024
Total views: 416.7k
Views today: 6.16k
Answer
VerifiedVerified
416.7k+ views
Hint: The sum of potential energy and kinetic energy at point A is equal to the kinetic energy at point D. The potential energy at point D is zero. The height in the formula of potential energy is 2h.

Complete step by step answer:
The roller coaster possesses both potential and kinetic energy at point A because potential energy $\left( {P_1} \right)$ is due to its height from the ground and kinetic energy $\left( {K_1} \right)$ due to its speed. A potential energy is the energy possessed by the body due to its specific position, configuration, etc and kinetic energy is energy that a body has due to its motion.
Thus, initial sum of total energy $ = {P_1} + {K_1}$
When the roller coaster reaches at point D, it only possesses kinetic energy. Potential energy is zero because its height from the ground is zero. Let the final kinetic energy be ${K_2}$. Thus, final sum of total energy ${K_2}$.
According to the law of conservation of energy, the energy can neither be created nor destroyed but it can be transferred from one form to another form of energy i.e., the initial sum of total energy is equal to the final sum of total energy.
${P_1} + {K_1} = {K_2}$
$Mg\left( {2h} \right) + \dfrac{1}{2}M{u^2} = \dfrac{1}{2}M{v^2}$ where M is the mass of roller coaster, g is acceleration due to gravity, 2h is height of the coaster at point A from the ground, and v is the final speed at point D.
$M\left( {2gh + \dfrac{1}{2}{u^2}} \right) = \dfrac{1}{2}M{v^2}$
$\Rightarrow 2gh + \dfrac{1}{2}{u^2} = \dfrac{1}{2}{v^2}$
$\Rightarrow \dfrac{{4gh + {u^2}}}{2} = \dfrac{{{v^2}}}{2}$
$\Rightarrow 4gh + {u^2} = {v^2}$
$\Rightarrow v = \sqrt {4gh + {u^2}} $

Therefore, option C is correct.

Note: The roller coaster is moving with uniform motion as the friction between the roller coaster and the track on which it is moving is negligible and it always remains in contact with it. Therefore, the total energy is conserved.