question attached A tennis ball of mass mt is held just above a basketball of ma

Question

question attached

A tennis ball of mass mt is held just above a basketball of mass mb, as shown in the figure below. 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 h 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? (Use any variable or symbol stated above along with the following as necessary: g.) ht = How do you account for the height in (a) being larger than h? Does that seem like a violation of conservation of energy?

Explanation / Answer

a). Since there was no initial velocity, you can use v = sqrt(2as) ........................... b) Because this is an elastic collision, we must use both the conservation of momentum and the conservation of energy to solve the problem. So, let u1 = before collision velocity of the tennis ball u2 = before collision velocity of the basketball v1 = after collision velocity of the tennis ball v2 = after collision velocity of the basketball m1 = mass of the tennis ball m2 = mass of the basketball .............................................. Then conservation of energy says m1u1^2/2 + m2u2^2/2 = m1v1^2/2 + m2v2^2/2................................. and conservation of momentum says m1u1 + m2u2 = m1v1 + m2v2.......................................... So the solution for v1 is v1 = (u1(m1 - m2) + 2m2u2) / (m1 + m2)......................... 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, 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 peak of it's bounce up................ g = acceleration due to gravity............... d = height the ball is at when it reaches the peak......................... So.............. (1/2)m1*v1^2 = m1*g*d................. (1/2)*v1^2 = g*d................. d = v1^2/(2*g)...................

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.