A tennis ball of mass 0.04kg is held just above a basketball of mass 0.5kg. With
ID: 1569614 • Letter: A
Question
A tennis ball of mass 0.04kg is held just above a basketball of mass 0.5kg. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height 1.5m and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling.
The two balls meet in an elastic collision. To what height does the tennis ball rebound?
Explanation / Answer
speed of basket ball u1 = sqrt(2*g*h)
mass of basket ball m1 = 0.5 kg
speed of tennis ball u2 = -sqrt(2*g*h)
mass of the tennis ball m2 = 0.04 kg
initial momentum before collision
Pi = m1*u1 + m2*u2
after collision final momentum
Pf = m1*v1 + m2*v2
from momentum conservation
total momentum is conserved
Pf = Pi
m1*u1 + m2*u2 = m1*v1 + m2*v2 .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2
KEf = 0.5*m1*v1^2 + 0.5*m2*v2^2
KEi = KEf
0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2 .....(2)
solving 1&2
we get
v1 = ((m1-m2)*u1 + (2*m2*u2))/(m1+m2)
v2 = ((m2-m1)*u2 + (2*m1*u1))/(m1+m2)
v2 = ( (-(0.04-0.5)*sqrt(2*g*h)) + (2*0.5*sqrt(2gh)) ) /(0.04+0.5)
v2 = 2.7*sqrt(2*g*h)
height rebounded = v2^2/2g = 10.935 m <<<<------answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.