91Ó°ÊÓ

1. Length of string,$L=2.0\text{m}$
2. Angle of string with vertical,$\mathrm{θ}=30°$
3. Velocity,$v=2.29m/s$
4. All the above values are taken as it is from the results of problem 7 of the same chapter.

Using the given values in problem 7 and energy conservation law, the velocity at the lowest point can be found. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae are as follow:

1. Energy conservation law,$E_{total}=\text{constant}.$
2. Total energy,$E_{total}=KE+PE$
3. $KE=\frac{1}{2}m{v}^2$
   
4. $PE=mgh$
   

where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, $E_{total}$is total energy and h is height.

From the geometry of the figure above,

$h=L-L\left(\cos \mathrm{θ}\right)$

$h=L\left(1-\left(\cos \mathrm{θ}\right)\right)$

At the lowest point, the ball possesses only kinetic energy,

$E_{total}=\text{constant}$

role="math" localid="1663140833799" $$\begin{gathered} K{E}_i+P{E}_i=K{E}_f+P{E}_f \\ 0+mgh=\frac{1}{2}m{v}^2+0 \\ v=\sqrt{2gh} \\ v=\sqrt{2gL\left(1-\left(\cos \mathrm{θ}\right)\right)} \end{gathered}$$

After plugging the values,

$v=\sqrt{2\left(9.8\right)\left(2.0\right)\left(1-cos30\right)}$

$v=2.29m/s$

Hence, speed of ball at the lowest point is$2.29m/s$.

As, the derived formula of velocity is,

$v=\sqrt{2gL\left(1-\left(\cos \mathrm{θ}\right)\right)}$

It can be seen that it doesn’t depend on themass, so for different masses, the value of velocity will be the same.

Hence, there is no effect of mass on speed.