You drop a 0.5kg ball from a height of 2m, and it bounces back to a height of 1.
ID: 1425016 • Letter: Y
Question
You drop a 0.5kg ball from a height of 2m, and it bounces back to a height of 1.5m. Consider a system formed by the ball and the Earth, so we can speak properly of its gravitational potential energy.
(a) What is the kinetic energy of the ball just before it hits the ground?
(b)What is the kinetic energy of the ball just after it bounces up?
(c) What is the coefficient of restitution for this collision?
(d) What kind of collision is this (elastic, inelastic, etc.)? Why?
(e) If the coefficient of restitution does not change, how high would the ball rise on a second bounce?
(f) On the graphs to the right, fill in the energy bar diagrams for the system: (1) as the ball leaves your hand; (2) just before it hits the ground; (3) just after it leaves the ground on its way up, and (4) at the top of its (first) bounce. Make sure to do this to scale, consistent with the values for the energies you have calculated above.
Explanation / Answer
here,
mass of the ball , m = 0.5 kg
height , h = 2 m
bounced height , h' = 1.5 m
(a)
the kinetic energy of the ball just before it hits the ground , KE = change in potential energy
KE = m * g * h = 0.5 * 9.8 * 2
KE = 9.8 J
(b)
the kinetic energy of the ball just after it bounces up, KE = m * g * h'
KE = 0.5 * 9.8 * 1.5
KE = 7.35 J
(c)
the coefficient of restitution for this collision , e = sqrt( KEafter impact / KEbefore impact)
e = sqrt( 7.35 /9.8)
e = 0.866
the coefficient of restitution for this collision is 0.866
d)
the kind of collison is inelastic
as the kinetic energy is not conserved during the collison
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