Question

The long pendulum shown drawn aside until the ball has risen 0.50 m. This
arrangement is placed on a planet where acceleration due to gravity is 7m/s2
It is then given an initial speed of 3.0m/s. The speed (in m/s) of the ball at its
lowest position is:
J
3.5m​

Answer 1

Given:

The long pendulum shown drawn aside until the ball has risen 0.50 m. This arrangement is placed on a planet where acceleration due to gravity is 7m/s². It is then given an initial speed of 3.0m/s.

To find:

Speed of the ball at the lowest position ?

Calculation:

In this kind of questions, we need to apply the principle of CONSERVATION OF MECHANICAL ENERGY.

The net change in potential energy will be equal to the change in kinetic energy.

• Let velocity at lowest point be v , initial velocity be u:

$\rm \therefore \: \Delta KE = - \Delta PE$

$\rm\implies \: \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {u}^{2} = - (0 - mgh)$

$\rm\implies \: \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {u}^{2} = mgh$

$\rm\implies \: \dfrac{1}{2} {v}^{2} - \dfrac{1}{2} {u}^{2} = gh$

$\rm\implies \: {v}^{2} - {u}^{2} = 2 gh$

$\rm\implies \: {v}^{2} - {(3)}^{2} = 2(7)(0.5)$

$\rm\implies \: {v}^{2} - 9 = 7$

$\rm\implies \: {v}^{2} = 16$

$\rm\implies \: v = \sqrt{16}$

$\rm\implies \: v = 4 \: m/s$

So, Velocity of ball at lowest position is 4 m/s.

Answer 2

The speed of the ball at its lowest point is 4 m/s

Explanation:

Given:

Value of acceleration due to gravity g = 7 m/s²

Height of the ball h = 0.5 m

Initial speed of the ball u = 3 m/s

To find out:

The speed of the ball in its lowest position

Solution:

At the lowest position the height = 0

Let the mass of the ball is m

The initial kinetic energy at height h

$K_i=\frac{1}{2}mu^2$

Initial Potential Energy at height h

$P_i=mgh$

If the velocity of the ball at its lowest point is v then

Final Kinetic energy of the ball

$K_f=\frac{1}{2}mv^2$

Final Potential Energy

$P_f=0$

Thus, by the conservation of the energy

$K_i+P_i=K_f+P_f$

$\implies \frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2+0$

$\implies u^2+2gh=v^2$

$\implies 3^2+2\times 7\times 0.5=v^2$

$\implies v^2=9+7$

$\implies v^2=16$

$\implies v=\sqrt{16}$

$\implies v=4$ m/s

Hope this answer is helpful.

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