# physics question

joe shmo
Posers and Puzzles 07 Mar '09 02:37
1. joe shmo
Strange Egg
07 Mar '09 02:371 edit
hello, I'm having a bit of difficulty solving the following problem...It is Physics 1 material..but seems to be tricky..

the jist of it

a spring gun, mounted firmly to a horizontal table top is loaded with a marble. The spring is compressed 1.0 cm and launched.( a horizontal launch, no vertical component of initial velocity) The marble falls 20 cm short of a target that is 2.0 m away from the base of the table, how far should the spring be compressed so it hits the target below?

spring constant = ?

mass = ?

initial velocity = ?

height = ?

im thinking to use a combination of work/energy, and standard kinematics...introducing another variable of time...but this seems to be really convoluted..

any pointers greatly appriciated.

Eric
2. 07 Mar '09 05:31
The marble traveled 1.8m, and you need to go 2m. Since time to hit the ground is constant, the ratio of the speed of the marble with the 1 cm compression to the speed you actually need is 9/10.

Then use the regular spring energy=kinetic energy equation and that should be pretty much it.

I get 1.111111....cm as the answer. I program all day so I might not remember physics and this might not be right ðŸ˜›. GO DISCLAIMERS!
3. coquette
07 Mar '09 07:42
Originally posted by Amarantine
The marble traveled 1.8m, and you need to go 2m. Since time to hit the ground is constant, the ratio of the speed of the marble with the 1 cm compression to the speed you actually need is 9/10.

Then use the regular spring energy=kinetic energy equation and that should be pretty much it.

I get 1.111111....cm as the answer. I program all day so I might not remember physics and this might not be right ðŸ˜›. GO DISCLAIMERS!
I don't think that the time to hit the ground is constant, but I think your approach is right on. Just put in the gravity formula for the time and that should work, right?
4. 07 Mar '09 12:50
The time to hit the ground is constant, because there is no vertical velocity. (Ignoring the fact the the path of an object is an ellipse no a parabola, such a short distance, you'll forgive me, right?)
5. coquette
07 Mar '09 17:35
Originally posted by Dejection
The time to hit the ground is constant, because there is no vertical velocity. (Ignoring the fact the the path of an object is an ellipse no a parabola, such a short distance, you'll forgive me, right?)
Oops. Sorry. You are right. I read it wrong.
6. PBE6
Bananarama
09 Mar '09 17:03
Originally posted by Amarantine
The marble traveled 1.8m, and you need to go 2m. Since time to hit the ground is constant, the ratio of the speed of the marble with the 1 cm compression to the speed you actually need is 9/10.

Then use the regular spring energy=kinetic energy equation and that should be pretty much it.

I get 1.111111....cm as the answer. I program all day so I might not remember physics and this might not be right ðŸ˜›. GO DISCLAIMERS!
I got a slightly different answer.

From Hooke's Law, we know that:

F = -kx

where:

"F" is force
"k" is the spring constant
"x" is the distance the spring has been stretched/compressed

From Newton's 2nd Law of motion, we know that:

F = ma

where:

"m" is the mass of the marble
"a" is the acceleration experienced by the marble

Equating the two and solving for "a", we have:

a = -kx/m

Note that "a" is expressed as a function of "x" here. Now, taking the integral of both sides with respect to "x", and choosing the limits of integration as x_i and x, we have:

INT(x_i:x)(a) = INT(x_i:x)(-kx/m)

The expression on the left is simply the velocity "v" as a function of "x". Re-writing and completing the integral on the right side, we have:

v(x) - v(x_i) = -k/(2m)*(x^2 - x_i^2)

Our coordinate system is such that the spring lies along the x-axis with the end of the table (i.e. the launch point) being the origin and the initial compressed state being less than 0 (i.e. x = -0.10 m for the original launch). Assuming the initial velocity v(x_i) to be 0, and letting x=0 to find the launch velocity, we have:

v(0) = k/(2m)*x_i^2

We now have an expression for the initial velocity as a function of the spring compression. Let's call this equation (1). Now a slight digression:

After the launch, the only force acting on the marble is the force of gravity, so the differential equations describing the marble's decent are:

d^2(y)/dt^2 = -g

d^2(x)/dt^2 = 0

Solving these equations with the following boundary conditions:

dy/dt(t=0) = 0
dx/dt(t=0) = v(0)
y(t=0) = h
x(t=0) = 0

We have:

y = -(1/2)*gt^2 + h
x = v(0)*t

Setting y = 0 and solving for "t", we have:

t = SQRT(2h/g)

Subbing this value into the equation for the motion in the "x" direction, we have:

x = v(0)*SQRT(2h/g)

We now have an expression for the distance traveled by the marble as a function of its initial velocity. Let's call this equation (2). Subbing equation (1) into equation (2), we get:

x = (x_i^2)*k/(2m)*SQRT(2h/g)

We now have an expression for the distance traveled by the marble as a function of its initial velocity. Since there are several constants in this equation that we don't know, and as it turns out we don't need to know, let's lump them all into a constant "C" and simplify:

x = C*(x_i^2)

Let's call this equation (3). Going back to the problem statement, we know that the marble traveled 1.8 m with a spring compression of 0.010 m. Subbing these values into equation (3), we have:

1.8 = C*(0.010^2) = C/10,000

Solving for C, we have:

C = 1.8*10,000 = 18,000

Now, we want to know how far we have to compress the spring in order to hit the target 2.0 m from the edge of the table. Subbing the distance into equation (3) and solving for x_i, we have:

2.0 = C*(x_i^2) = 18,000*(x_i^2)

x_i = SQRT(2.0/18,000) = 0.01054 m = 0.011 m = 1.1 cm

Therefore, you will need to compress the spring by an extra tenth of a centimetre to make the target. Amarantine's estimate of 1.1 cm is a good one, and would work in this case (and provide the same answer when rounded to the correct number of significant digits), but it gets less accurate the further off the original shot was. For instance, if the original shot was short by a full 1.0 m, the estimate would give you an answer of 2.0 cm, while the calculation above give you an answer of x_i = SQRT(2.0/10,000) = 1.4 cm.

Interesting problem!
7. uzless
The So Fist
09 Mar '09 20:473 edits
Originally posted by PBE6
I got a slightly different answer.

From Hooke's Law, we know that:

F = -kx

where:

"F" is force
"k" is the spring constant
"x" is the distance the spring has been stretched/compressed

From Newton's 2nd Law of motion, we know that:

F = ma

where:

"m" is the mass of the marble
"a" is the acceleration experienced by the marble

Equating f x_i = SQRT(2.0/10,000) = 1.4 cm.

Interesting problem!
This explanation is like using a nuke to kill an ant!

ðŸ˜‰
8. uzless
The So Fist
09 Mar '09 20:55
Originally posted by joe shmo
hello, I'm having a bit of difficulty solving the following problem...It is Physics 1 material..but seems to be tricky..

the jist of it

a spring gun, mounted firmly to a horizontal table top is loaded with a marble. The spring is compressed 1.0 cm and launched.( a horizontal launch, no vertical component of initial velocity) The marble falls 20 cm sh ...[text shortened]... of time...but this seems to be really convoluted..

any pointers greatly appriciated.

Eric
Pointers? Keep this simple...don't overcomplicate things.

The distance to target is 200 centimeters. Shot went 180cm.

In other words, 10% short. So, compress the spring an extra 10%.

10% of 1cm= 1.1cm.
9. uzless
The So Fist
09 Mar '09 20:571 edit
Originally posted by uzless
10. 09 Mar '09 22:231 edit
EDIT: Following message to PBE6 (and I apologize in advanced if I sound arrogant....I talk like a textbook, I've been told).

The units for your initial velocity don't work out.

As you stated,

a = -kx/m

and

v(0) = k/(2m)*x_i^2

However, in terms of units, "v(0)" and "a" differ only by a factor of meters...so something isn't right.

Also, you can find the initial velocity by equating spring energy with kinetic energy

kx^2/2= mv^2/2 => v= x*sqrt(k/m)

I think where it went wrong was in taking the integral. You can solve it without taking integrals of anything. I'm not sure what's wrong with the integral. I'll look at it more closely and post if I find anything.
11. 09 Mar '09 22:27
I think I found it.

INT(x_i:x)(a) = INT(x_i:x)(-kx/m)

The expression on the left is simply the velocity "v" as a function of "x". Re-writing and completing the integral on the right side, we have:

v(x) - v(x_i) = -k/(2m)*(x^2 - x_i^2)

INT(x_i:x)(a) is NOT velocity as a function of x. It would've be something like INT(t_i:t)(a). Incidentally, this is where the units go wrong (m^2/s * m turned into m/s here).
12. 09 Mar '09 22:35
Originally posted by uzless
Pointers? Keep this simple...don't overcomplicate things.
The distance to target is 200 centimeters. Shot went 180cm.
In other words, 10% short. So, compress the spring an extra 10%.
10% of 1cm= 1.1cm.
I think Amarantine's and uzless' solutions work perfectly.
Why complicate things when they don't need to be complicated?
13. PBE6
Bananarama
09 Mar '09 23:23
Originally posted by Amarantine
I think I found it.

INT(x_i:x)(a) = INT(x_i:x)(-kx/m)

The expression on the left is simply the velocity "v" as a function of "x". Re-writing and completing the integral on the right side, we have:

v(x) - v(x_i) = -k/(2m)*(x^2 - x_i^2)

INT(x_i:x)(a) is NOT velocity as a function of x. It would've be something like INT(t_i:t)(a). Incidentally, this is where the units go wrong (m^2/s * m turned into m/s here).
Aha, right you are.
14. joe shmo
Strange Egg
10 Mar '09 01:13
Originally posted by PBE6
Aha, right you are.
just a question, Even though your derivation turned out to be invalid due to units, would you use the same approach to find a solution, or would you just use the porportions as the others...?
15. PBE6
Bananarama
10 Mar '09 12:40
Originally posted by joe shmo
just a question, Even though your derivation turned out to be invalid due to units, would you use the same approach to find a solution, or would you just use the porportions as the others...?
I prefer to understand a problem well enough that I can derive the answer from first principles, so yes, if asked this question again (for the first time!) I would use the same approach but with the correct equation. I find that I learn much more this way - just look at this thread! Concision and clarity are the ultimate goal of course, but the detailed step-by-step solution can be an illuminating first draft.