Solve Spring & Ball Puzzle: Find m1, m2, k & Sm1

In summary, the ball falls off the platform and its energy is shared between the two balls. The ball on the left has a smaller energy and falls faster.
  • #1
Peethassandwich
2
0
1. You have one platform (1.2m high and 0.12m wide) and a few small balls of a different mass (0.09, 0.14, 0.25, 0.38, 0.57, 0.9 kg) that slide across the platform without friction and you have 4 springs with various stiffness (556, 918, 1110, 1546 N/m, ignore the mass of the spring). You can lay 2 balls on a platform (the heavier one on the left and the easier one on the right) and lay arbitrary spring on the platform squeezed by 0.02 m (the legth in its natural state is 0.1m). You have to (by using the spring) to make the balls fall from the platform onto the two buttons that are 2 metres apart from each other. If you want the ball to press the button it has to land max. 0,007m away from its centre. What is the value of Sm1, m1, m2 and k?
https://imageshack.com/i/eyledVFlp

I am really confused by this task so forgive me if you find some blunder in my solution
The first thing I am not sure about is whether the ball was still accelerating when it fell off the platform.
And since the force exerted on this ball behaved in a kind of "dirac delta" way I was wondering whether I could get this involved in calculating initial velocity of the ball. That is really the only thing I can't really figure out. I projected this motion into x-y plane and set up the following equations (let's say for the ball on the left):

X-axis sm1 = ? v(initial) = ? t = ?
Y-axis g = 9.81 m/s^2 d = 1.2 m v(initial) = 0 m/s
Then I used this equation : d = v(initial)*t + 1/2*g*t^2
so : 1.2 = 0.5*9.81*t^2
t = 0.494619 s
So according to this the ball was falling for 0.494619 s.
Then I tried to apply formula kX=ma but in this particular case I couldn't somehow get this thing involved since k,m,a are unknown.
X = 0.02m
F = k*X
F = 1/50*k
Thanks in advance for your help.

Edit : I posted an image as a reply since I couldn't display it in the problem.
 
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  • #2
springs.png
 
  • #3
The first problem you need to solve is how the energy of the spring will be shared out between the two balls. What conservation laws can you think of that might apply? Remember that the spring is not fixed to the platform, and you are to ignore its mass.
 

FAQ: Solve Spring & Ball Puzzle: Find m1, m2, k & Sm1

1. What is the purpose of solving the Spring & Ball Puzzle?

The purpose of solving the Spring & Ball Puzzle is to determine the values of the mass of the ball (m1), the mass attached to the spring (m2), the spring constant (k), and the displacement of the mass attached to the spring (Sm1).

2. How do you solve the Spring & Ball Puzzle?

The Spring & Ball Puzzle can be solved by using the equation F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement of the mass attached to the spring. By measuring the displacement of the mass and the force applied by the spring, the values of m1, m2, k, and Sm1 can be calculated.

3. What are the units of measurement for the variables in the Spring & Ball Puzzle?

The units of measurement for the variables in the Spring & Ball Puzzle are: mass (m) in kilograms (kg), spring constant (k) in newtons per meter (N/m), displacement (x) in meters (m), and force (F) in newtons (N).

4. Can the Spring & Ball Puzzle be solved with different units of measurement?

Yes, the Spring & Ball Puzzle can be solved with different units of measurement as long as they are all consistent. For example, if the mass (m) is measured in grams, then the force (F) should be measured in newtons, and the displacement (x) should be measured in centimeters.

5. What are the applications of solving the Spring & Ball Puzzle?

Solving the Spring & Ball Puzzle has various applications in physics and engineering, such as understanding the behavior of springs and systems with oscillating masses, designing and optimizing mechanical systems, and calculating the forces and displacements in structures subjected to vibrations.

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