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Elastic collision with a spring constant and unknown masses
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[QUOTE="HallsofIvy, post: 5384199, member: 637751"] 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 [itex]v= \sqrt{1.5g}[/itex] so its momentum is [itex]m\sqrt{1.5g}[/itex]. On the level, both "conservation or kinetic energy" and "conversation of momentum" hold. [/QUOTE]
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Elastic collision with a spring constant and unknown masses
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