Physics, 13.11.2019 03:31, tlgbulldog38

Apackage of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiting truck. unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

a. what is the speed of the package of mass m right before the collision?

b. suppose the packages stick together. what is their common speed after the collision? is the total mechanical energy for the system including both packages conserved in this case? if not, what is the difference of the mechanical energies before and after the collision?

c. suppose the collision between the packages is perfectly elastic. to what height does the package of mass m rebound? is the total linear momentum for the system including both packages conserved before and after this perfectly elastic collision? why or why not?

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answered: mlandon922

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where is Kinetic energy and is Potential energy.

$K= \frac{mv^2}{2}$ and U= mgh

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.Sum of KE and PE is also known as Mechanical energy.Only KE is conserved for elastic collision.for elastic collison we have e=1

that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with 2.56 then there is a -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

answered: Guest

Centrapedial force is the answer

answered: Guest

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

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Apackage of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high fr...