A tennis ball with a mass of 57.0 g is held just above a basketball with a mass of 580 g. With their centers aligned vertically, both balls are released from rest at the same time to fall a distance of 1.00m.
a) You should not round your answer to more than 3 significant digits as this is the best you can get with the data for this problem (possibly up to 2 significant digits, depending on the accuracy of this value for the mass of the basketball ). Since there was no initial velocity, you can use v = sqrt(2as) = sqrt(2 * 9.807 m/s^2 * 1 m) = 4.428769581 m/s = 4, 43 m/s like you did.
b) Since this is an elastic collision, we need to use both conservation of momentum and conservation of energy to solve the problem. (Note that the first answer is incorrect, as it incorrectly assumes that the velocity after the basketball collision is negligible.) Then let
u1 = velocity before tennis ball collision = -4.43 m/s
u2 = before basketball collision velocity = 4.43 m/s
v1 = collision speed after tennis ball
v2 = collision velocity after basketball
m1 = mass of the tennis ball = 57.0 g
m2 = mass of the basketball = 589 g
So the conservation of energy says
m1u1^2/2 + m2u2^2/2 = m1v1^2/2 + m2v2^2/2
and the conservation of momentum says
m1u1 + m2u2 = m1v1 + m2v2
The only two true unknowns in these two equations are v1 and v2 and with two equations you can solve them. However, the solution is long and confusing (I’ve done one of these types of problems before today), so I’ll cheat a bit and give the answer. (Note, if you need to show all your work, a good tip is to change the reference frame so that one of the initial velocities is zero, then convert back once you have the solution – see my source.)
So the solution for v1 is
v1 = (u1(m1 – m2) + 2m2u2) / (m1 + m2)
v1 = ((-4.429m/s)(57.0g – 580g) + 2(580g)(4.429m/s)) / (57.0g + 580g)
v1 = ((-4.429m/s)(-523g) + (1160g)(4.429m/s))/(637g)
v1 = ((2316m/s) + (5138m/s))/637
v1 = (7454 m/s)/637 = 11.7 m/s
(FYI, v2 is about 2.84 m/s, and you can plug those numbers back into the two conservation equations and you’ll see they’re correct.)
Anyway, now we have an upward velocity for the tennis ball. The kinetic energy for this will be fully converted into potential energy when the tennis ball reaches its maximum height, so
KE = (1/2)m1*v1^2 = PE = F*d = m1*g*d
Where
KE = Kinetic energy of the tennis ball immediately after the collision
PE = Potential energy of the tennis ball at the top of its upward bounce
g = acceleration due to gravity
d = height at which the ball is when it reaches the top
So
(1/2)m1*v1^2 = m1*g*d
(1/2)*v1^2 = g*d
d = v1^2/(2*g)
d = (11.7m/s)^2/(2 * 9.807m/s^2)
d = 6.98m
(You can also use the standard kinematic equations to get this distance, the result is the same.)
Not only is the number not given by you, but you forgot to ask the question. Please ask the question again. Thanks