Solving GRE Spheres: Find h0 from h

In summary, the conversation discusses an exercise involving two spheres of putty, A and B, with masses M and 3M respectively, hanging from the ceiling on strings of equal length l. Sphere A is raised to a height h0 and then released, colliding with sphere B and sticking together. The conversation mentions using the conservation of energy to find the maximum height h they swing to, but the correct answer involves considering momentum conservation due to the inelastic collision.
  • #1
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"Two small spheres of putty, A and B, of mass M and 3M, respectively, hang from the ceiling on strings of equal length l. Sphere A is drawn aside so that it is raised to a height h0 and then released. Sphere A collides with sphere B; they stick together and swing to a maximum height h equal to "

This is an exercise I am perplexed about as to why I got the wrong answer. I used conservation of energy.

Ei = mgh0

Ef = (m+3m)gh

Ef = Ei

mgh0 = (m+3m)gh

Therefore h = (1/4)m

This answer is wrong. The correct answer is (1/16)h0. Where'd I go wrong? Thanks!
 
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  • #2
They stick together after collision, that (as well as the fact that they are made of putty) should tell you that collision is inelastic and kinetic energy is not conserved. You have to consider momentum conservation to calculate Ef.
 
  • #3
Thanks!
 

1. How do I solve for h0 in GRE spheres?

To solve for h0 in GRE spheres, you can use the formula h0 = h/((R^2-h^2)^0.5), where h is the height of the sphere and R is the radius of the sphere. This formula can be derived using the Pythagorean theorem and trigonometric ratios.

2. What is the importance of solving for h0 in GRE spheres?

Solving for h0 in GRE spheres is important because it allows you to calculate the distance between the center of the sphere and a point on its surface. This information is crucial in various fields such as physics, engineering, and mathematics.

3. Can I use the same formula to solve for h0 in different types of spheres?

Yes, the formula h0 = h/((R^2-h^2)^0.5) can be used to solve for h0 in any type of sphere, whether it is a perfect sphere or an oblate or prolate spheroid. The only variables that may change are the values of h and R.

4. Are there any other methods to solve for h0 in GRE spheres?

Yes, there are other methods to solve for h0 in GRE spheres, such as using the Law of Cosines or the Law of Sines. These methods involve using the angles and side lengths of the triangle formed by the center of the sphere, the point on its surface, and the point on the x-axis.

5. Can I use the formula to solve for h0 if h is greater than R?

No, the formula h0 = h/((R^2-h^2)^0.5) is only applicable when h is less than R. If h is greater than R, the point on the surface of the sphere would be outside the sphere and the distance between the center and the point cannot be calculated using this formula.

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