91Ó°ÊÓ

The mass of larger ball is,$M=0.63\mathrm{kg}$.

The two balls are dropped simultaneously from the height $h=1.8\mathrm{m}$.

By usingtheconservation of mechanical energy and by finding thevelocity of the ball of massMand massmafter collision, find thevalue ofthat result in the larger ball stopping when it collides with the small ball and theheight that the small ball reaches.

Formulae are as follow:

The velocity of the ball of massMafter collision is,$v_{Mf}=\frac{M-m}{M+m}{v}_{Mi}+\frac{2m}{M+m}{v}_{mi}$

The velocity of the ball of massafter collision is,role="math" localid="1661312732417" $v_{mf}=-\left(\frac{m-M}{m+M}\right)\sqrt{2gh}+\frac{2M}{M+m}\sqrt{2gh}$

The velocity is,$v_{Mi}=v_{mi}=\sqrt{2gh}$

The conservation of mechanical energy is,$Mgh=\frac{1}{2}M{v}^2$

Here, the initial kinetic energy is zero, the initial gravitational potential energy is Mgh, the final kinetic energy is $\frac{1}{2}M{v}^2$and the final potential energy is zero. Thus, the conservation of mechanical energy is given by,

$Mgh=\frac{1}{2}M{v}^2$

And,

$v=\sqrt{2gh}$

The collision of the ball ofM with the floor is an elastic collision of the light object that reversesthe direction without change in magnitude. After the collision, the ball is travelling upward with speed$v=\sqrt{2gh}$. The ball of massis traveling downward withthe same speed.

Hence, the velocity of the ball of mass M after collision is,

$$\begin{gathered} v_{Mf}=\frac{M-m}{M+m}{v}_{Mi}+\frac{2m}{M+m}{v}_{mi} \\ {v}_{Mf}=\frac{M-m}{M+m}\sqrt{2gh}-\frac{2m}{M+m}\sqrt{2gh} \\ {v}_{Mf}=\frac{M-3m}{M+m}\sqrt{2gh} \end{gathered}$$

For this to be zero,$m=\frac{M}{3}$,

$$\begin{gathered} m=\frac{M}{3} \\ m=\frac{0.63}{3} \\ m=0.21\mathrm{kg} \end{gathered}$$

Hence,the value of results in the larger ball stopping when it collides with the small ball is, m=0.21 kg.

Similarly, the velocity of the ball of massmafter the collision is given by,

$$\begin{gathered} v_{Mf}=\left(\frac{m-M}{m+M}\right)\sqrt{2gh}+\frac{2m}{M+m}\sqrt{2gh} \\ {v}_{mf}=\frac{3M-m}{M+m}\sqrt{2gh} \end{gathered}$$

Substituting M=3m,

$v_{mf}=2\sqrt{2gh}$

Now, the initial kinetic energy is,$\frac{1}{2}m{v}_{mf}^2$, initial potential energy is zero, final kinetic energy is zero and final potential energy is mgh. Thus, by applying conservation of mechanical energy,

$$\begin{gathered} \frac{1}{2}m{v}_{mf}^2=mgh\text{'} \\ h\text{'}=\frac{v_{mf}^2}{2g} \\ h\text{'}=\frac{4×2gh}{2g} \\ h\text{'}=4h \\ h\text{'}=4×1.8 \\ h\text{'}=7.2m \end{gathered}$$

Hence,the small ball reaches a height of$h\text{'}=7.2\mathrm{m}$.

Therefore, by using the conservation of mechanical energy and by finding the velocity of the balls after collision, the height of the ball can be found.