Question

A ball of mass m shown in figure, slides on a semi-
circular frictionless track of radius R. If it starts from
rest at position A what is its speed at the point marked
B?​

Answer 1

Given:

A ball of mass m shown in figure, slides on a semi-circular frictionless track of radius R. It starts from rest at position A.

To find:

Speed at point B ?

Calculation:

In these of questions, we will use the principle of CONSERVATION OF MECHANICAL ENERGY.

• So, the change in potential energy will be equal to the change in the kinetic energy as per the principle.

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

$\implies \: mg \bigg\{l \cos( \theta) \bigg \} = \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {u}^{2}$

$\implies \: mg \bigg\{l \cos( \theta) \bigg \} = \dfrac{1}{2} m {v}^{2} - \dfrac{1}{2} m {(0)}^{2}$

$\implies \: mg \bigg\{l \cos( \theta) \bigg \} = \dfrac{1}{2} m {v}^{2}$

$\implies \: g \bigg\{l \cos( \theta) \bigg \} = \dfrac{1}{2} {v}^{2}$

$\implies \: {v}^{2} = 2g \bigg\{l \cos( \theta) \bigg \}$

$\implies \: v = \sqrt{2g \bigg\{l \cos( \theta) \bigg \} }$

$\implies \: v = \sqrt{2 \times 10 \times 1 \times \cos( {45}^{ \circ} ) }$

$\implies \: v = \sqrt{2 \times 10 \times 1 \times \frac{1}{ \sqrt{2} } }$

$\implies \: v = \sqrt{ \sqrt{2} \times 10 }$

$\implies \: v = \sqrt{ 1.414 \times 10 }$

$\implies \: v = \sqrt{ 14.14 }$

$\implies \: v = 3.76 \: m/s$

So, speed at Point B is 3.76 m/s.