A tennis ball of mass 57.0 g is held just above a basketballof mass 590 g. With

Question

A tennis ball of mass 57.0 g is held just above a basketballof mass 590 g. With their centers vertically aligned, bothballs are released from rest at the same time, to fall through adistance of 2.00 m.

a) Find the magnitude of the downward velocity with which thebasketball reaches the ground.

b) Assume that an elastic collision with the ground instantaneouslyreverses the velocity of the basketball while the tennis ball isstill moving down. Next, the two balls meet in an elasticcollision. To what height does the tennis ballrebound?
A tennis ball of mass 57.0 g is held just above a basketballof mass 590 g. With their centers vertically aligned, bothballs are released from rest at the same time, to fall through adistance of 2.00 m.

a) Find the magnitude of the downward velocity with which thebasketball reaches the ground.

b) Assume that an elastic collision with the ground instantaneouslyreverses the velocity of the basketball while the tennis ball isstill moving down. Next, the two balls meet in an elasticcollision. To what height does the tennis ballrebound?

Explanation / Answer

For part a) use v2=u2+2aS (where S=2m, u=0because the balls are at rest and a=9.8m/s2--gravity) v=6.26m/s b) since the collision is elastic, the total momentum of the system isconserved. Since mass is not a factor in determining the final velocity of theballs (and assuming the balls are point masses, otherwise theradius of each would change the height through which they fall),both the final velocities are 6.26 m/s downwards. By conservation of momentum: initial momentum=final momentum (ofthe system, where momentum (p)=m*v) Now the basketball's momentum is reversed and is pointing upwardsbecause of the rebounding. and is equal to -6.26m/s. Note that energy is also conserved so the two equations we getare: m1u1+m2u2=m1v1+m2v2 and 1/2m1u12+1/2m2u22=1/2m1v12+1/2m2v22 makingv1 the subject of the formula we get: v1=u1*(m1-m2)/(m1+m2)+ 2m2*u2/(m1+m2) v1=-5.157-11.42 =-16.58m/s (minus sign means velocity is directed upwards) Now we use v2=u2+2aS again where v=0 since wewant the height at which the ball stops.. Solve for S and we get ... S=14m

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