To conserve both momentum and energy, the first ball stops and the second one continues with the total energy.
In part 2, momentum is conserved but the motion continues with twice the mass.
m g (.29)(1-cos 25) = .5 m v^2
solve for v
then conservation of momentum at bottom
(m+m) * new v = m v
new v = (1/2) v
then at top
g (.29)(1-cos A) = .5 (new v)^2
solve for A
Two identical steel balls, each of mass 5.0 kg, are suspended from strings of length 29 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle θ = 25° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?
Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision?
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2 answers