91影视

We have given,

height of the slider =$3\mathrm{m}$

We have to find the common speed of the package after the collision.

firstly we will find the speed of the mass m before collision.

Then, we can find it using conservation of energy that change in kinetic energy will be equal to the change in potential energy.

$鈭�\mathrm{K}.\mathrm{E}=鈭�\mathrm{P}.\mathrm{E}$

The change in kinetic angry of the mass m is given by,

$$\begin{gathered} 鈭�\mathrm{K}.\mathrm{E}.=\mathrm{kineic}\mathrm{energy}\mathrm{just}\mathrm{before}\mathrm{the}\mathrm{collosion}-\mathrm{kinetic}\mathrm{enrgy}\mathrm{at}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{inclined} \\ 鈭�\mathrm{K}.\mathrm{E}.=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}\mathrm{m}脳0 \\ 鈭�\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2......\left(1\right) \end{gathered}$$

Then, change in potential energy will be due to change in the height of the mass m is

$$\begin{gathered} 鈭�\mathrm{P}.\mathrm{E}.=\mathrm{mg}鈭�\mathrm{h} \\ 鈭�\mathrm{P}.\mathrm{E}.=(3\mathrm{m})(9.8\mathrm{m}/{\mathrm{s}}^2)\mathrm{m} \\ 鈭�\mathrm{P}.\mathrm{E}.=29.4\mathrm{m}.....\left(2\right) \end{gathered}$$

Using equation (1) and (2)

$$\begin{gathered} \frac{1}{2}m{v}^2=29.4m \\ {v}^2=58.8 \\ v=7.67m/s \end{gathered}$$

This is the speed of the mass m just before the collision. then using the conservation of energy.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}} \\ \left(\mathrm{m}\right)(7.67\mathrm{m}/\mathrm{s})=(2\mathrm{m}+\mathrm{m})\left(\mathrm{v}\right) \\ \mathrm{v}=\frac{7.67}{3} \\ \mathrm{v}=2.56\mathrm{m}/\mathrm{s} \end{gathered}$$

We have given,

height of the slider=$3\mathrm{m}$

We have to find the at what height the mass m will rebound.

In the elastic collision the energy and momentum of the system will be conserved. then,

From the conservation of momentum,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{i}}={\mathrm{P}}_{\mathrm{f}} \\ \mathrm{m}(7.67\mathrm{m}/\mathrm{s})=\mathrm{m}\left({\mathrm{v}}_1\right)+2\mathrm{m}({\mathrm{v}}_{2)} \\ 7.67={\mathrm{v}}_1+2{\mathrm{v}}_2 \\ {\mathrm{v}}_1=7.67-2{\mathrm{v}}_2.......(3) \end{gathered}$$

From the conservation of energy,

$$\begin{gathered} \frac{1}{2}m{(7.67m/s)}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}\left(2m\right)v_2^2 \\ 58.8=v_1^2+2v_2^2 \end{gathered}$$

Now putting the value of velocity from the equation (3) in this equation.

$$\begin{gathered} 58.8={\mathrm{v}}_1^2+\frac{1}{2}{(7.67-{\mathrm{v}}_1)}^{2}58.8=\frac{3}{2}{\mathrm{v}}_1^2+29.4-7.67{\mathrm{v}}_1 \\ {\mathrm{v}}_1=2.56\mathrm{m}/\mathrm{s} \end{gathered}$$

Then using the conservation of energy change in kinetic energy is equal to change in potential energy,

$$\begin{gathered} \frac{{(2.56)}^2}{2}=(9.8m/s^2)\left(h\right) \\ h=0.33m. \end{gathered}$$