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Verdich [7]
3 years ago
11

A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho

ut rolling). The slope makes an angle θ with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb.
Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

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Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

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                      = 2 \times 0.1 m

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = \frac{k \times Q \times Q}{R}

                    = \frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}

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As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, \frac{U}{2}.

Hence,   K.E = \frac{U}{2}

                     = \frac{0.081}{2}

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Now, we will calculate the speed of balls as follows.

                 V = \sqrt{\sqrt{\frac{2 \times K.E}{m}}

                     = \sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

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