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Jakob Littel

Feb 20, 2025

A tennis ball of mass 57.0 g is held just above a basketball of mass 580 g.?

A tennis ball of mass 57.0 g is held just above a basketball of mass 580 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.00 mNeed Help with Part B(a) Find the magnitude of the downward velocity with which the basketball reaches the ground.I got that to be 4.427 m/sAssume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. (b) To what height does the tennis ball rebound?

2 Answers

H
Hermina Ferry

Feb 20, 2025

a) You should round your answer to at most 3 significant digits because that is the best you can get with the data in this problem (possibly even only 2 significant digits, depending on how precise that value is 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 as you basically have.b) Because this is an elastic collision, we must use both the conservation of momentum and the conservation of energy to solve the problem. (Note the first answer is incorrect as it incorrectly assumes that the after collision velocity of the basketball is negligible.) So, letu1 = before collision velocity of the tennis ball = -4.43 m/s u2 = before collision velocity of the basketball = 4.43 m/sv1 = after collision velocity of the tennis ballv2 = after collision velocity of the basketballm1 = mass of the tennis ball = 57.0 gm2 = mass of the basketball = 589 gThen conservation of energy saysm1u1^2/2 + m2u2^2/2 = m1v1^2/2 + m2v2^2/2and conservation of momentum saysm1u1 + m2u2 = m1v1 + m2v2The 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 messy (I already did one of these types of problems earlier today), so I am going to cheat a little and give you the answer. (Note, if you need to show all your work, a good tip is to change the frame of reference so that one of the initial velocities is zero and then convert back after you have the solution – see my source.)So the solution for v1 isv1 = (u1(m1 – m2) + 2m2u2) / (m1 + m2)v1 = ((-4.429 m/s)(57.0 g – 580 g) + 2(580 g)(4.429 m/s)) / (57.0 g + 580 g)v1 = ((-4.429 m/s)(-523 g) + (1160 g)(4.429 m/s))/(637 g)v1 = ((2316 m/s) + (5138 m/s))/637v1 = (7454 m/s)/637 = 11.7 m/s(FYI, v2 is about 2.84 m/s, and you can plug these figures back into the two conservation equations and you will see that they are correct.)Anyway, now we have an upwards velocity for the tennis ball. The kinetic energy for for this will be converted entirely into potential energy when the tennis ball reaches maximum height, soKE = (1/2)m1*v1^2 = PE = F*d = m1*g*dWhereKE = Kinetic Energy of the tennis ball immediately after the collisionPE = Potential Energy of the tennis ball at the peak of it’s bounce upg = acceleration due to gravityd = height the ball is at when it reaches the peakSo (1/2)m1*v1^2 = m1*g*d(1/2)*v1^2 = g*dd = v1^2/(2*g)d = (11.7 m/s)^2/(2 * 9.807 m/s^2)d = 6.98 m(You can also use the standard kinematic equations to get this distance, the result is the same.)... Show More
H
Hermina Ferry

Feb 20, 2025

Not only the figure is not given by you but you have forgotten to ask the question. Please ask the question again. Thanks... Show More

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