A mass m is placed at the rim of a frictionless hemispherical bowl with a radius R and released from
rest as in (Figure 1). It then slides down and undergoes a
perfectly elastic collision with a second mass 3m sitting
at rest at the bottom of the bowl.

A. What is the speed of the first mass just after the collision?

B. What is the speed of the second after the collision?

C. To what maximum height will the first travel after collision.

(All answers in terms of g, m,R)

Explanation:

at collision, m has a velocity of √2gR

The center of mass of the system has a velocity of

m(√2gR) + 3m(0) = (m + 3m)v

v = ¼√2gR

The stationary mass (3m) will depart at twice the system center of mass velocity

v(3m) = ½√2gR   ANSWER B

The relative velocity before the collision will be the negative of the relative velocity after the collision.

v(m) = ½√2gR - √2gR = -½√2gh ANSWER A

m will rise to a height of v²/2g = (-½√2gR)²/2g = 0.25R

where is the photo and the answer choices?

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2