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 3msitting at rest at the bottom of the bowl.

Figure 1:https://session.masteringphysics.com/problemAsset/2036997/4/YF-08-045.jpg

What is the direction of the first mass just after the collision?

What is the direction of the second mass just after the collision?

To what maximum height will the first mass travel after the collision? (Express your answer in terms of R.)

Explanation / Answer

let potential energy referrence point be at the bottom of the bowl.

==>at the bottom of the bowl, potential energy=0

now, initialy the mass m is at height R.

then its potential energy=m*g*R

as its speed=0, its kinetic energy=0

hence total mechanical energy=m*g*R+0=m*g*R

when it reaches the bottom, just before the collison:

height=0==>potential energy=0

using conservation of energy principle,

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)

now, momentum of the system before collison=m*v+3*m*0=m*v

let after collison, mass m moves to the left with a speed of v1 and mass 3m moves to the right with a speed of v2.

then using conservation of momentum principle and assuming the initial direction of momentum (i.e. to the right) to be +ve

initial momentum of the system=final momentum of the system

==>m*v=3*m*v2-m*v1 (there is a negative sign for momentum of mass m because it is moving to the left i.e. opposite to the direction which we are considering as positive)

==>3*v2-v1=v

==>v1=3*v2-v...(1)

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

using equation 1 for v1,

(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

using this value of v2 in equation 1,

we get v1=3*v2-v

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

hence 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)

when it reaches its maximum height, its speed will be 0.

hence if it reaches maximum height of h,

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

==>h=0.125*v^2/g

using v=sqrt(2*g*R)

we get,

maximum height=0.25*R

final answers are:

a)direction of first mass just after collison is left.

b) direction of second mass just after collision is right.

c) maximum height reached by mass m is R/4.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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