Simple harmonic motion pinball problem

In summary, the conversation discusses a problem involving a ball being shot from a spring in a pinball machine. The spring has a constant of 675 N/m and is compressed by 0.0650m. The ball has a mass of 0.0585kg and is initially at rest against the spring at point A. When released, the ball slides to point B, which is 0.300m higher than point A. The conversation then discusses how to calculate the speed of the ball at point B, using conservation of energy. It is determined that the final velocity can be found by setting the final energy equal to the initial energy, and solving for the final velocity. After some confusion and calculation, the final velocity is determined
  • #1
I have been looking at this problem forever. It is probably easy, but I am making it harder. States: In preparation for shooting a ball in a pinball machine, a spring (k=675 N/m) is compressed by 0.0650m relative to its unstrained length. The ball ( m=0.0585kg) is at rest against the spring at point A. When the spring is released, the ball slides (w/o rolling) to point B, which is 0.300m higher then point A. How fast is the ball moving in B?

you want to find the vf (velocity final), I think so I set up the equation as 1/2vf2 + 1/2 kxf2= 1/2 kxo2 (o=initial, f=final)
vf=sqrt km(xo2-xf2) which I got an outrageous number for. Please help I am soo lost
 
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  • #2
Do it using conservation of energy.

1) Calculate the potential energy stored in the spring when strechted. Once released, all that energy becomes kinetic.

2) Calculate what energy the ball has converted in gravitational potential once it reaches B.

3) The remains of the energy is still kinetic. Find the velocity associated to that amount of kinetic energy.
 
  • #3
So you would set up the equation as E= 1/2MV2 + mgh + 1/2kx2 right
 
  • #4
Yeah, you rock! Then use the fact that [itex]E_f = E_0[/itex] to recover [itex]v_f[/itex].
 
  • #5
what do u mean to recover vf
 
  • #6
so 1/2(.058kg)(0 m/s) + 1/2(.058kg)(9.80 m/s2)(h?) + 1/2(675 N/m)(x?)2
so xo=.0650 and then xf is .300 m higher so xf=.3650m right
 
  • #7
lol, what? Be consistent with your notation. You got 2 values of xf there.
What you got to do is set Ef = Eo, plug all the values, solve for vf.
 
  • #8
no the problem says that point B is .300m higher then a so I thinmk that would mean that xf would = .3650m bc u would add the beginning xo value plus the .300m value. Also I don't know what h would be and xfor the equation
 
  • #9
so we would use 1/2mvf2 + 1/2 kxf2= 1/2mvo2 +1/2 kvo2 from the equation Ef =Eo
 
  • #10
h is 0.300 and the Xo of the spring's pot.energy is 0.0650m.

The problem is 2 dimensional. When they say point B is 0.300m higher, it means higher in the sense of 'h'.
 
  • #11
heather32283 said:
so we would use 1/2mvf2 + 1/2 kxf2= 1/2mvo2 +1/2 kvo2 from the equation Ef =Eo
No you're missing the gravitationa potential energy now. You had it right in your earlier post.
 
  • #12
/2mvf2 + mgho +1/2 kxf2= 1/2mvo2 + mghf +1/2 kvo2 right
 
  • #13
heather32283 said:
1/2mvf2 + mgho +1/2 kxf2= 1/2mvo2 + mghf +1/2 kxo2 right

Ef = Eo means (all final values) = (all initial values). Not a mix of both. Keep it to

1/2mvf2 + mghf +1/2 kxf2= 1/2mvo2 + mgho +1/2 kxo2
 
  • #14
opps had it right on paper but put it in the wrong area on here. So would xf=.650 +.300, that is the only variable I am trying to find, other then the final speed
 
  • #15
Xf is 0. It represent the length by which the spring is stretched! As soon as the ball leaves the spring x is 0.
 
  • #16
so hf=.300, ho= .0650 ??
 
  • #17
ho = 0

xo = 0.0650
 
  • #18
so we plug in the numbers and solve for vf right
 
  • #19
right...
 
  • #20
that will then give us the speed at B, god you have been such a help thank you
 
  • #21
does 27.4 m/s sound right
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a system moves back and forth around a central equilibrium point, with a restoring force proportional to the displacement from the equilibrium position.

2. How does a pinball machine demonstrate simple harmonic motion?

In a pinball machine, the ball moves back and forth in a pendulum-like motion as it bounces off the pins and bumpers. This motion is caused by the force of gravity pulling the ball down and the elastic force of the bumpers pushing it back up, creating a simple harmonic motion.

3. What factors affect the simple harmonic motion in a pinball machine?

The factors that affect simple harmonic motion in a pinball machine include the gravitational force, the elasticity of the bumpers and pins, and the initial velocity and direction of the ball.

4. How can the equations of motion be used to analyze a simple harmonic motion pinball problem?

The equations of motion, such as the equation for position, velocity, and acceleration, can be used to calculate the position, velocity, and acceleration of the ball at any given time during its motion. This information can then be used to analyze the behavior of the ball in the pinball machine.

5. Can the simple harmonic motion of a pinball machine be affected by external factors?

Yes, the simple harmonic motion of a pinball machine can be affected by external factors such as air resistance, friction, and the shape and layout of the machine. These factors can alter the motion of the ball and make it deviate from a perfect simple harmonic motion.

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