Examkrackers physics Q.67

[obog360]

---

Members do not see this ad.

Question: Objects A and B are placed on the spring. Object A has twice as much mass as object b. If the spring is depressed and released propelling the objects into the air, Object A will:

answer given: C. rise to the same height as object B.

I thought it should have been B. rise half as high as object B.

my reasoning: all of spring energy is converted to kinetic energy and then to gravitational potential energy. The spring is compressed the same distance for both A and B (because they are on the spring together)
1/2kx^2 = mgh
because B=m and A=2m so A will rise to 1/2 height of B.

can anyone tell me where my reasoning is wrong?
thanks

 

[marele86]

Because in projectile motion, mass does not matter.

As the spring moves back towards equilibrium with an acceleration a, the masses will move with the same acceleration. Why? Because they don't LEAVE the spring until the spring is back to equilibrium, and then, since they both had the same acceleration over the same period of time, they will have the same velocity.

Two objects of different mass, moving upwards with the same velocity is kinematics. y=v(o)t - 1/2at^2. Same acceleration, some initial velocity.. same height before they start coming down.

As far as the conservation of energy goes (which is what I'm guessing you really wanted to know, since EK explains the above, but not why you can't use CoE for the problem): what you write would be correct, EXCEPT that we have an outside force acting on our system. Initially the spring was depressed a certain distance, and it was NOT due to the masses (or else the spring would have just stayed down). That means we had someone pushing down on the spring. This is an outside force! It gave energy to our system (depressing a spring over a certain distance = work = energy), and an outside force (which is nonconservative) means that our system's energy will not be conserved.

It would, of course, be conserved if we viewed this outside force (the person pressing down on the spring) as part of our system, but we have no idea what force they used (we don't know the spring constant, so we can't use F=-kx) or the energy, so therefore we can't calculate the energy. Therefore kinematics is the way to go.

Hope that helps.

 

[physics junkie]

To echo Marlee,

Both objects receive the same acceleration but the heavier mass receives more force than the lighter mass receives. When the spring is back to its equilibrium length both masses are traveling at the same velocity but with different amounts of energy. If you think about it its quite impossible for one object to exit the spring before the other object does--they have to leave the spring at the same time.

Marlee is incorrect about conservation of energy though. Conservation of energy can be easily applied to this problem. Since both objects are leaving the spring with the same velocity you can consider it one object with mass 3m. This will come naturally if you just look at the fully written equation:
(1/2)kx^2 = (1/2)mv^2 + (1/2)(2m)v^2
(1/2)kx^2 = (1/2)(3m)v^2

You can therefore see why it is possible to use this equation to compute the height:
1/2kx^2 = 3mgh

But getting to this point means you already conceptually grasped the question that you posed.

Once again, (1/2)*mv^2 and (1/2)*2m*v^2 as the two kinetic energies of the light and heavy mass upon release. The heavier mass receives 2/3rds of the spring's kinetic energy and the lighter mass receives 1/3rd of the spring's kinetic energy.

Best,
Physics Junkie P.I.P.

 

[marele86]

To echo Marlee,

Marlee is incorrect about conservation of energy though. Conservation of energy can be easily applied to this problem. Since both objects are leaving the spring with the same velocity you can consider it one object with mass 3m. This will come naturally if you just look at the fully written equation:
(1/2)kx^2 = (1/2)mv^2 + (1/2)(2m)v^2
(1/2)kx^2 = (1/2)(3m)v^2

You can therefore see why it is possible to use this equation to compute the height:
1/2kx^2 = 3mgh

Thanks for the correction. 👍 Honestly not sure why I didn't think of it like that.. for some reason it just seemed so much harder. Nice!

 

[donkeykongsnes]

I actually have a question about the explanation given for the exam kracker practice test 1-H. I don't understand how they got the answer for questions 48 and 50. If someone could explain those questions to me I would greatly appreciate it.