Question

A package of mass m is released from rest at a warehouse loading dock and slides down the $3.0\mathrm{m}$-high, frictionless chute of FIGURE EX11.25 to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass $2\mathrm{m}$, from the bottom of the chute.

a. Suppose the packages stick together. What is their common speed after the collision?

b. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Short answer

a. Speed is $2.56\mathrm{m}/\mathrm{s}.$

b. The height is$0.33\mathrm{m}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

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

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

Simplify

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.

$∆\mathrm{K}.\mathrm{E}=∆\mathrm{P}.\mathrm{E}$

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

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

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

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

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.

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

Part (b) Step 1: Given information

We have given,

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

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

Simplify

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,

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