(a) In Problem 4, what initial speed must be given the ball so that it reaches the vertically upward position with zero speed? What then is its speed at

(b) The lowest point and

(c)The point on the right at which the ball is level with the initial point?

(d) If the ball’s mass were doubled, would the answers to (a) through (c) increase, decrease, or remain the same?

(i) Mass of ball is $M=0.341\text{kg}$,

(ii) Length of string is $L=0.452\text{m}$,

The problem is based on the law of conservation of energy which states that the total energy of an isolated system remains constant.According to the law of energy conservation, energy can neither be created, nor be destroyed.Using energy conservation and the concept of changing energy as per the position of the ball, find the velocities at various points as asked in the problem.

Formulae:

Kinetic energy is given by,

$KE=\frac{1}{2}m{v}^2$

Potential energy is given by,

$PE=mgh$

where,m is mass, v is velocity, g is an acceleration due to gravity andh is height.

As, the marble goes to the highest point change in the potential energy is,

$\Delta U=mgL$

In this case, “L” is considered as “h”

And, by the law of conservation,

$\Delta KE+\Delta PE=0$

$K_{top}-K_0+mgL=0$

At the top point kinetic energy is zero,

$$\begin{gathered} K_{top}=0 \\ {K}_0=mgL \end{gathered}$$

role="math" localid="1662975400172" $\begin{array}{rcl}\frac{1}{2}m{v}_0^2& =& mgL\\ v_0& =& \sqrt{2gL}\\ & =& \sqrt{2\left(9.8\right)\left(0.452\right)}\\ & =& 2.98\text{m/s}\end{array}$

Hence, initial speed given to the ball so that it reaches the vertical upward position with zero speed is$2.98\mathrm{m}/\mathrm{s}$

As, the object comes downwards, the change in potential energy is,

$\Delta U=-mgL$

And, by the law of conservation,

localid="1662975619403" $\begin{array}{rcl}\Delta KE+\Delta PE& =& 0\\ K_{bottom}-K_0-mgL& =& 0\end{array}$

And, from part (a)

$K_0=mgL$

$K_{bottom}-mgL-mgL=0$

$K_{bottom}=2mgL$

$\begin{array}{rcl}\frac{1}{2}m{v}_{bottom}^2& =& 2mgL\\ v_{bottom}& =& \sqrt{4gL}\\ & =& \sqrt{4\left(9.8\right)\left(0.452\right)}\\ & =& 4.21\text{m/s}\end{array}$

Hence, its speed at lowest point is$4.21\text{m/s}$

As, there is no change in height as in part a, so no change in potential energy. And also, no change in kinetic energy also. So, answer will same as part a.

$\begin{array}{rcl}{v}_{right}& =& v_0\\ & =& 2.98\text{m/s}\end{array}$

Hence, its speed atpoint on the right at which the ball is level with initial point is$2.98\text{m/s}$

As, all the derived formula, the velocity does not depend on the mass, so the velocity will be the same.

Hence, if the ball’s mass were doubled then there will be no changes in the answers of part (a) to part (c).