A block of mass m starts from rest from the edge of a hemispherical bowl of radi

Question

A block of mass m starts from rest from the edge of a hemispherical bowl of radius R and slides down the frictionless surface. when the block gets to its lowest point it collides and sticks to another block of mass 3m. Given (m,R) Determine:

a. the maximum angle the blocks slide up the other side of the bowl.

b. what fraction of the original energy was lost in the collision?

c. Redo the problem, except this time the blocks collide with perfect elasticity. Determine the maximum angle of each block after the collision.

Explanation / Answer

initial energy of mass m = m*g*R

at the bottom kinetic energy of the m = 0.5*m*u^2

from energy conservation

energy at top = energy at bottom

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

u = sqrt(2*g*R)

after collsion the two block stick together and move with a speed V

momentum before collision = momentum after collision

m*u = (m + 3m)*V

v = u/4 = sqrt(2*g*R)/4 = sqrt(gR/8) m/s

(a)

let theta be angle made with vertical

at theta angle the potential energy = PE = (m+3m)*g*R*(1-costheta))

after collision KE = 0.5*(m+3m)*v^2 = 0.5*4m*u^2/16 = 0.5*4*m*2*g*R/16 = m*g*R/4

from energy conservation

KE = PE

m*g*R/4 = (m+3m)*g*R*(1-costheta))

1/4 = 4*(1-costheta)

theta = 20.36 degrees <<_---------answer

(b)

KE loss = KEi - KEf = 0.5*m*u^2 - 0.5*(m+3m)*v^2

KE loss = 0.5*m*2*g*R - 0.5*4m*gR/8

KE loss = mgR - mgR/4 = 3mgR/4     <<<<<<<<----------answer

---

(c)

let v1 & v2 be the speed of m & 3m after collision

from momentum conservation

Pi = Pf

m*u = m*v1 + 3m*v2

u = v1 + 3v2

======> v2 = u-v1/3.............(1)

from energy conservation

0.5*m*u^2 = 0.5*m*v1^2 + 0.5*3m*v2^2 ...........(2)

from 1 & 2

v1 = u/2

v2 = u/2

let theta1 be angle made with vertical

for m

at theta1 angle the potential energy = PE = m*g*R*(1-costheta))

after collision KE = 0.5*m*v1^2 = 0.5*m*u^2/4 = 0.5*m*2*g*R/4 = m*g*R/4

from energy conservation

KE = PE

m*g*R/4 = m*g*R*(1-costheta1))

1/4 = (1-costheta1)

theta1 = 41.4 degrees <<_---------answer

for 3m

let theta2 be angle made with vertical

at theta2 angle the potential energy = PE = 3m*g*R*(1-costheta2))

after collision KE = 0.5*3m*v1^2 = 0.5*3m*u^2/4 = 0.5*3m*2*g*R/4 = 3*m*g*R/4

from energy conservation

KE = PE

3*m*g*R/4 = 3m*g*R*(1-costheta2))

1/4 = (1-costheta2)

theta2 = 41.4 degrees <<_---------answer

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