Catapult spring, Kinetic and Potential energy

[User1265]

Homework Statement

A catapult consists of a massless cup attached to a massless spring of length
l and of spring constant k.

i) If a ball of mass m is loaded into the cup and the
catapult pulled back to extend the spring to a total length x, what velocity
does the ball reach when launched horizontally?

ii) The catapult is then used to launch the ball vertically. If the spring is
extended to the same total length of x before release, to what velocity does
the catapult now accelerate the ball?


Q: Why does h = x - l

Relevant Equations

1/2kx^2
1/2 mv^2
mgh

My solutions: When ball is launched horizontally, assuming its velocity is entirely in the horizontal dimension, there is no interaction of the ball with the gravitational field, thus no change in GPE, so all of the EPE (elastic potential energy ) of the spring is transferred to KE of the ball.

(1/2) mv^2= 1/2 (k(x-l)^2)

i) v = √ (k/m) * (x-l)ii) When launched vertiically, now the mass is interacting with gravtiational field, as well as possessing maxmium kinetic energy (thus speed) at the moment of release.

so, total energy of system (ball) = KE +GPE = (1/2) mv^2 + mgh

1/2 (k(x-l)^2) = (1/2) mv^2 + mgh

Q1) But the solution mentions that h = x-l , why is this? :

With the diagram I have drawn ( attached), I cannot understand why h= x-l

I'm assuming the height from which it is released intially is zero. But does the ball have to gain a potential energy of the height it is displaced by (x-l) ? I don't quite understand why this.
1/2 (k(x-l)^2) = (1/2) mv^2 + mg(x-l)

when gives v velocity as the maximum velocity reached. It is assuming the greatest velocity reached by the ball is the instant it leaves the catapult, then it must travel a vertical distance of x-l to do so, and thus, gains Mg(x-l) as well as KE. But I don't understand why the maximum velocity is when the ball leaves the catapult.

 

[haruspex]

possessing maxmium kinetic energy (thus speed) at the moment of release.

Depends what you mean by the moment of release. What forces act on the mass when the spring is relaxed?

 

[User1265]

Depends what you mean by the moment of release. What forces act on the mass when the spring is relaxed?

The moment the ball leaves the catapult being the moment of release. Gravity acts on the mass when the spring is relaxed I believe.
Update:
The model the solution uses for the catapult is the mass being on the end of the catapult, which differs to the one I have drawn, as the mass would be pulled to extend the spring, whereas I thought the spring itself would be pulled.

 

[PeroK]

The moment the ball leaves the catapult being the moment of release. Gravity acts on the mass when the spring is relaxed I believe.
Update:
The model the solution uses for the catapult is the mass being on the end of the catapult, which differs to the one I have drawn, as the mass would be pulled to extend the spring, whereas I thought the spring itself would be pulled.

You have a choice in these problems:

1) Take the starting point to be the natural equilibrium point of the spring. You put the mass on the spring and push it down to a displacement $x_0$, say.

2) You put the mass on the spring first and take the starting point as the equilibrium with the spring supporting the mass. Then, you push the mass down by a displacement $x'_0 < x_0$.

You should be able to do it both ways, but you need to be careful about GPE.

 

[haruspex]

The moment the ball leaves the catapult being the moment of release.

Which would be when the spring is length L, right?

Gravity acts on the mass when the spring is relaxed I believe.

Yes, so will its upward speed then be a maximum or increasing or reducing?

Your diagram doesn't seem to make sense for sending the ball upwards (which I assume is the intended interpretation). Either the spring should be compressed to length x or the top of the spring is fixed and the mass is in a cup at the lower end.

 

[User1265]

Which would be when the spring is length L, right?

Yes

 

[User1265]

Yes, so will its upward speed then be a maximum or increasing or reducing?

At the point the ball leaves the spring I believe the upwards speed would be reducing due to the force of gravity on the ball.
But what I don't understand is why h=x-l ,
why is the gravitational potential energy mg(x-l) when ball leaves the spring?

 

[PeroK]

At the point the ball leaves the spring I believe the upwards speed would be reducing due to the force of gravity on the ball.
But what I don't understand is why h=x-l ,
why is the gravitational potential energy mg(x-l) when ball leaves the spring?

How are $x$ and $L$ defined?

I think you are right about the point of launch. There must be a variable there that depends on how quickly the spring loses contact with the projectile. If the spring is stiff, then the difference between the simple launch point (when the spring no longer supports the weight of the projectile) and the actual launch point (when the projectile actually loses contact with the spring) must be small.

I'm not sure you were supposed to worry about that.

 

[PeroK]

But what I don't understand is why h=x-l ,
why is the gravitational potential energy mg(x-l) when ball leaves the spring?

I assume that $x$ is the total compression of the sping and $L$ is the compression of the spring under the mass of the projectile. The compression point $L$ is considered the launch point. The GPE gained from the maximum compression to the launch point is $mg(x-L)$.

Note that an alternative approach is to take the equilibrium point at $L$ to begin with. Then the compression is $x - L$. The spring force in this case is reduced by the gravitational force and these forces and energies cancel out. In this case, therefore, you can consider the reduced PE of the spring only and ignore GPE.

You might like to check that the two calculations give the same result.

 

[User1265]

How are $x$ and $L$ defined?

I think you are right about the point of launch. There must be a variable there that depends on how quickly the spring loses contact with the projectile. If the spring is stiff, then the difference between the simple launch point (when the spring no longer supports the weight of the projectile) and the actual launch point (when the projecile actually loses contact with the spring) must be small.

I'm not sure you were supposed to worry about that.

X is the total length of the spring after compression. L is the natural length. With regards to the latter question, is my reasoning correct for the point of maxium velocity:
when ball loses contact with the spring, it's velocity is a maxium for the reason, when the spring is extended down by a extension of X-L and released, the spring's restoring force acts on the ball doing work over a distance of X-L on the ball, (the force decrealine linearly /gradually to zero at the equilibrium- when spring returns to ornignal length L ), and so the spring accelerates the ball up until the point zero contact, and elastic potential energy is transferred to ball is 1/2 (x-l)^2. Velocity at this point must be maximum, as the net acceleration after this point ( i.e in freefall) is in the opposite direction due to gravity.

 

[haruspex]

At the point the ball leaves the spring I believe the upwards speed would be reducing due to the force of gravity on the ball.

So it has already lost some speed, right? The question asks
"to what velocity does the catapult now accelerate the ball?"
I take that to mean the maximum speed reached.