Elastic collision with a spring constant and unknown masses

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SUMMARY

The discussion revolves around solving a physics problem involving an elastic collision between two balls of different masses and the compression of a spring. The first ball, with mass m, rolls down a 3.0 m ramp inclined at 30°, colliding elastically with a second ball of mass 2m, which is initially at rest. The spring constant is given as k = 200 N/m, and the spring compresses by 0.15 m. The key equations utilized include conservation of energy and momentum, leading to the determination of the mass m through the relationship between kinetic energy and spring potential energy.

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  • Understanding of elastic collisions in physics
  • Knowledge of conservation of momentum and energy principles
  • Familiarity with spring mechanics and Hooke's Law
  • Basic trigonometry for resolving forces on an incline
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  • Study the principles of elastic collisions in detail
  • Learn about conservation of momentum and energy in closed systems
  • Explore the calculations involved in spring compression and potential energy
  • Review the application of trigonometric functions in physics problems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of elastic collisions and energy conservation principles.

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Homework Statement


A ball of mass m rolls down a 3.0 m ramp inclined at 30° above the horizontal, rolls along a flat, friction less surface, and collides elastically with another ball of mass 2m, initially at rest. The second mass then moves along the surface and collides with a horizontally mounted spring with spring constant k= 200 N/m, compressing it by 0.15 m. What is the value of m?

Homework Equations


Ek 1 + Eg 1 + Ee 1 = Ek 2 + Eg 1 + Ee 2

For the incline:
FN + Fg parallel + Fg perpendicular + Ff = m1a

The Attempt at a Solution


I've been working backwards so I started with this:

1/2 m2v2' 2 = 1/2 kx2

I ended up with v2' = √45/2m and I now have no idea how to continue

Any help will be appreciated.
 
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The top of the 3 m long, 30 degree ramp is 3 sin(30)= 1.5 m above its base. Taking the kinetic energy of the first ball to be 0 at the top of the ramp, its potential energy is 1.5mg, relative to the bottom. At the bottom of the ramp, it potential energy is 0 so its kinetic energy is 1.5mg and its velocity is given by v= \sqrt{1.5g} so its momentum is m\sqrt{1.5g}. On the level, both "conservation or kinetic energy" and "conversation of momentum" hold.
 
Oops. I forgot about conservation of momentum. Thanks!
 

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