A mass m is placed at the rim of a frictionless hemispherical bowl with a radius

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 3m sitting at rest at the bottom of the bowl. To what maximum height will the first mass travel after the collision? Express your answer in terms of R. To what maximum height will the second mass travel after the collision? Express your answer in terms of R. What is the speed of the second mass just after the collision? Express your answer in terms of g and R.What is the speed of the first mass just after the collision? Express your answer in terms of g and R.

Explanation / Answer

at the bottom of the bowl, potential energy=0

initialy the mass m is at height R.

then its potential energy=m*g*R = total mechanical energy

final kinetic energy+final potential energy=total mechanical energy of the system

0.5*m*v^2+0=m*g*R

0.5*m*v^2=m*g*R

v=sqrt(2*g*R)

--------------------------------------------------------------------------------------

Apply cosnervation of mometum to the mases

m*v+3*m*0=3*m*v2-m*v1

3*v2-v1=v

v1=3*v2-v.

using conservation of energy principle for just before collison and just after collison:

0.5*m*v^2+0.5*3*m*0^2=0.5*m*v1^2+0.5*3*m*v2^2

v1^2+3*v2^2=v^2

---------------------------------------------------------------------------------------

(3*v2-v)^2+3*v2^2=v^2

9*v2^2+v^2-6*v*v2+3*v2^2=v^2

12*v2^2=6*v*v2

v2=0.5*v

since v1=3*v2-v

v1=3*0.5*v-v=0.5*v

--------------------------------------------------------------------------------------------

the mass m will travel with a speed of 0.5*v to the left after collison

then total energy of mass m just after collison=0.5*m*(0.5*v)^2=0.125*m*v^2 (potential energy=0)

if it reaches maximum height of h,

m*g*h=0.125*m*v^2

h=0.125*v^2/g

subsitute v=sqrt(2*g*R) in the above equaiton

H_max =0.25*R

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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