Linear Relationship b/w Spring Compression and Projectile Distance?

In summary: I'm just wondering then how come the distance is not linear. Can't we say that V is linearly proportional to the spring force, and then use that to show that distance is linearly proportional to the spring force?In summary, the conversation discussed a physics exercise involving a spring gun and a marble. The question was whether the horizontal distance traveled by the marble and the amount the spring is compressed have a linear relationship. The conversation explored the forces at play, such as the spring force and gravity, and used equations to determine that the speed of the marble is proportional to the spring force, but the distance traveled by the marble is not linearly proportional to the spring force.
  • #1
Steven60
7
1
I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework.
Thanks!
 
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  • #2
Seems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this?
 
  • #3
This is a nice physics exercise - there are several physical considerations, and then some simple math.

You have two forces acting on the marble ... the spring force, which launches the marble, and gravity.

Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.

The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.

Then the distance from the table to the point of contact will be D = V x T.

The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.

Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.

If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.

But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.

Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.

Thus the hypothesis is true!

Thanks to dauto for noticing the mistake at the end! :-)
 
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  • #4
The equations involved will be ##d=vt## for constant ##v## (and assuming that the initial point when exiting the spring gun is defined as 0 distance), ##U_{spring}=\frac12k\Delta x^2## and ##K=\frac12mv^2##. ##\Delta x## is the amount the spring is compressed, and ##v## is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for ##v## to find the relationship between ##\Delta x## and ##d##.
 
  • #5
UltrafastPED said:
This is a nice physics exercise - there are several physical considerations, and then some simple math.

You have two forces acting on the marble ... the spring force, which launches the marble, and gravity.

Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.

The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.

Then the distance from the table to the point of contact will be D = V x T.

The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.

Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.

If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.

But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.

Thus V = k/M Sqrt[X].

Thus the hypothesis is false - the distance covered by the marble is NOT a linear function of the compression distance for the spring.

You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.
 
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  • #6
Thanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.
 

Related to Linear Relationship b/w Spring Compression and Projectile Distance?

What is a linear relationship?

A linear relationship is a type of mathematical relationship between two variables where the data points form a straight line when plotted on a graph. This means that as one variable increases or decreases, the other variable also changes at a constant rate.

How does spring compression affect projectile distance?

Spring compression refers to the amount that a spring is compressed or stretched from its resting position. In terms of projectile distance, the more a spring is compressed, the greater the force it exerts on the projectile, resulting in a longer distance travelled by the projectile.

What factors can affect the linear relationship between spring compression and projectile distance?

There are several factors that can affect the linear relationship between spring compression and projectile distance. These may include the mass of the projectile, the initial velocity of the projectile, and the force applied to compress the spring. Other external factors such as air resistance may also play a role.

How can the linear relationship between spring compression and projectile distance be demonstrated experimentally?

To demonstrate the linear relationship between spring compression and projectile distance, one could set up an experiment where a projectile is launched using a spring-loaded launcher. By varying the amount of spring compression and measuring the distance travelled by the projectile each time, a data set can be obtained and plotted on a graph. If the data points form a straight line, it would confirm the linear relationship between the two variables.

Why is it important to understand the linear relationship between spring compression and projectile distance?

Understanding the linear relationship between spring compression and projectile distance is important as it allows us to predict the distance a projectile will travel based on the amount of spring compression. This knowledge can be applied in fields such as engineering and physics, where the use of springs and projectiles are common.

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