Question

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 3msitting at rest at the bottom of the bowl. Part A: What is the speed of the first mass just after the collision? Part B: What is the direction of the first mass just after the collision? Part C: What is the speed of the second mass just after the collision? Part D: What is the direction of the second mass just after the collision? Part E: To what maximum height will the first mass travel after the collision Part F: To what maximum height will the second mass travel after the collision?

Answer 1

finding the initial speed of the first mass, at the rim all energy is potential and at the botton all energy is kinetic, so:

$mgR = \frac{1}{2}mv_{1i}^2 \Rightarrow v_{1i}= \sqrt{2gR}$

momentum conservation:

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$\rightarrow m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$\rightarrow mv_{1i} = mv_{1f} + 3mv_{2f}$

$\rightarrow v_{1i} = v_{1f} + 3v_{2f}\;\;\;[1]$

kinetic energy conservation (linear form)

$v_{1i} - v_{2i} = -(v_{1f}-v_{2f})$

$\rightarrow v_{1i} = -(v_{1f}-v_{2f})$

$\rightarrow v_{1i} = -v_{1f}+v_{2f}\;\;\;[2]$

equating equations [1] and [2]

$v_{1f} = -v_{2f}$

therefore:

$v_{1f} = -\frac{1}{2}v_{1i} = -\frac 12 \sqrt{2gR}$

and

$v_{2f} = \frac{1}{2}v_{1i} = \frac 12 \sqrt{2gR}$

part a)

$v_{1f} = -\frac{1}{2}v_{1i} = -\frac 12 \sqrt{2gR}$

part b)

left

part c)

$v_{2f} = \frac{1}{2}v_{1i} = \frac 12 \sqrt{2gR}$

part d)

right

I can help you with the other parts but you need post again as a new question