Is it possible to apply energy conservation here?

[navneet9431]

Homework Statement

Homework Equations

Kinetic Energy =1/2*m*v^2
Spring Potential Energy = 1/2*k*x^2
Gravitational Potential Energy = m*g*h

The Attempt at a Solution

I am thinking to solve this problem using energy conservation but I feel that it is not possible to apply energy conservation because there is an external force of the spring on the block at the moment the block B collides with block A.Isn't it?
I will be thankful for help!

 

[PKM]

I am thinking to solve this problem using energy conservation but I feel that it is not possible to apply energy conservation

Why not? Energy conservation can be applied to the spring-&-two-mass system. Block B sticks to block A after collision, so what about considering the two blocks jointly as a single one?

 

[navneet9431]

Why not? Energy conservation can be applied to the spring-&-two-mass system. Block B sticks to block A after collision, so what about considering the two blocks jointly as a single one?

Thanks!
I got it.
One more doubt,
At what level shall i consider the gravitational potential energy zero?

 

[haruspex]

I am thinking to solve this problem using energy conservation but I feel that it is not possible to apply energy conservation because there is an external force of the spring on the block at the moment the block B collides with block A.

The external force of the spring is not a bar to using work conservation, but the impact is. You can only apply that law from after the impact.

 

[haruspex]

At what level shall i consider the gravitational potential energy zero?

Potentials are always relative. You can set the zero wherever you like here. In some other contexts, there are standard conventions.

 

[PKM]

At what level shall i consider the gravitational potential energy zero

To elaborate @haruspex's post, for example, gravitational P.E. is considered zero at Earth surface, just as for the problem you posed. For the mass in this problem, it is $mgh$ at height $h$ from Earth surface.
But actually the gravitational P.E. of an object due to a gravitating body is exactly zero at infinite distance from the body.
See more at: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/gravpe.html

 

[PeroK]

To elaborate @haruspex's post, for example, gravitational P.E. is considered zero at Earth surface, just as for the problem you posed. For the mass in this problem, it is $mgh$ at height $h$ from Earth surface.
But actually the gravitational P.E. of a point due to a gravitating body is exactly zero at infinite distance from the body.
See more at: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/gravpe.html

Considering GPE to be 0 at infinity is equally arbitrary. Note also that in that case 0 is the maximum GPE and it is negative everywhere else.

 

[haruspex]

For the mass in this problem, it is mgh at height h from Earth surface.

That is unknown, so not convenient.
Take it as zero at the initial position of A.

 

[Delta2]

Another mini problem I see with this type of problems, is that the sum of external forces won't remain equal to zero during the collision. We have to assume that the collision is instantaneous in time and that it starts and ends at the same position (the initial position of the mass at A), so that we can apply conservation of momentum along the vertical axis.

 

[BvU]

I got it.

Can you show ?

 

[haruspex]

Another mini problem I see with this type of problems, is that the sum of external forces won't remain equal to zero during the collision. We have to assume that the collision is instantaneous in time and that it starts and ends at the same position (the initial position of the mass at A), so that we can apply conservation of momentum along the vertical axis.

It doesn't bother me so much here. If the collision takes time dt then error in conservation of momentum will be of order dt².

 

[PKM]

That is unknown, so not convenient.
Take it as zero at the initial position of A.

Yes, you're right. What I meant here is that it can be taken as $mgh$ depending on height $h$; I used it to illustrate my point only.
In this problem, the difference between the PE's all that matters.

 

[kuruman]

To add my two-bits, I will argue that energy is indeed conserved in this case. Not conserved are momentum and mechanical energy. Total energy at point A just before the moving masses collide is equal to total energy at point B, where the combined masses rise to the point where the spring is at its "natural length". One can write the total energy conservation equation as $\Delta K+\Delta U+\Delta E_{dis.}=0.$ The first two terms have their usual meanings. The last term is the energy change due to the dissipative forces during the collision and is a positive quantity because the internal energy of the blocks rises. It's always the same when masses stick together as long as no external forces accelerate the CM while the sticking takes place (referred to in post #9). In that case it is equal to the kinetic energy in the CM frame of reference:$$\Delta E_{dis.}=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2=\frac{1}{2}\mu v_{rel}^2$$where $\mu$ is the reduced mass and $v_{rel}$ is the relative speed before the collision.

An application of this is the ballistic pendulum, where point A is the collision point and point B is maximum height reached by the two masses. We have $\Delta K = 0 - \frac{1}{2}m_1v_0^2$, $\Delta U=(m_1+m_2)gh-0$. Then$$- \frac{1}{2}m_1v_0^2+(m_1+m_2)gh+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}v_0^2=0$$This can easily be solved to find $v_0$ in terms of the other quantities. Total energy is always conserved; one just has to find how much is hiding where when energy transformations take place.