Spring cannon, initial velocity, and compression

In summary: I used the conservation of energy equation: (potential energy of spring)= (kinetic energy of balloon) + (heat energy lost) so (1/2)(350)(x^2) = (1/2)(0.5)(v^2) + 5 solving for v gives v = 15.04 m/sIn summary, the problem involves launching a 0.5 kg water balloon with a spring cannon at a 30 degree angle and a distance of 20m. To achieve this, the initial velocity needs to be 15.04 m/s and the spring needs to be compressed by 0.56 m. However, there may be a discrepancy in the values as the calculation for the initial velocity
  • #1
fischelr
1
0

Homework Statement


Part a.) You launch a 0.5 kg water balloon with a spring cannon in order to his someone 20m away. The spring in the cannon has a spring constant of 350 N/m. The spring cannon fires the balloon at a 30 degree angle with respect to the ground. What is the initial velocity necessary for the balloon to travel 20 meters?

Part b.) How much does the spring need to be compressed in order to achieve this initial velocity? Assume that 5 joules of heat is lost as the spring expands.


Homework Equations



Fnet = Fspring - lwl = ma
sohcahtoa
*something for energy loss*

The Attempt at a Solution



Part a.) V initial was found to be 14.09 m/s

Part b.)
First I determined the lwl parallel be 30sin 4.9=2.45N

Fnet = Fspring - lwl = ma
=k[tex]\Delta[/tex]x - lwl = ma
=350 x - 2.45 = 2.45
350x = 4.90
x=.014m
x-1.4cm


I have found that the spring compression would need to be 1.4cm to launch a .5kg ball 20 meters. However, this sounds small to me. Am I missing a force on the spring? I thought about including power, but I do not have the acceleration of the ball leaving the spring.

Also, I am not sure how to subtract 5 joules of energy due to heat loss.
 
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  • #2
For part a, I did not get 14.09 m/s, can you post how you got that answer?

For part b, you could use conservation of energy here. But resultant force or 'ma' would not be 2.45 N, you'd need to find the acceleration.
 
  • #3
On part B would it not simply be the potential energy of the spring = to the kinetic energy plus heat?

so .5(350)x^2 = .5(.5)14.09^2 + 5

x^2 = 54.6/175

x= .56 m
 
  • #4
Yes that would be correct to do.
 
  • #5
would a velocity of 15.04 be correct for this?
 
  • #6
Yes that should be the initial velocity.
 
  • #7
what equation did you use to get a velocity of 15.04? Thanks!
 
  • #8
stephjuly28 said:
what equation did you use to get a velocity of 15.04? Thanks!

The better question is what did you use to get 14.09 m/s?
 

Related to Spring cannon, initial velocity, and compression

1. What is a spring cannon?

A spring cannon is a device that uses the energy stored in a compressed spring to launch a projectile or object.

2. How is the initial velocity of a spring cannon calculated?

The initial velocity of a spring cannon can be calculated using the equation v = √(kx^2)/m, where v is the initial velocity, k is the spring constant, x is the distance the spring is compressed, and m is the mass of the projectile.

3. What factors affect the initial velocity of a spring cannon?

The initial velocity of a spring cannon is affected by the spring constant, distance the spring is compressed, and the mass of the projectile. Other factors that may affect the initial velocity include friction, air resistance, and the angle at which the projectile is launched.

4. How does compression affect the performance of a spring cannon?

The amount of compression in a spring cannon affects the initial velocity and therefore the range of the projectile. The more the spring is compressed, the greater the potential energy stored in the spring and the higher the initial velocity of the projectile.

5. Can a spring cannon be used for scientific experiments?

Yes, a spring cannon can be used for scientific experiments to study projectile motion, potential and kinetic energy, and the effects of different variables on the performance of the cannon.

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