The question is #2 Ball A (2 lbs) is released from rest and subsequently slides

Question

The question is #2
Ball A (2 lbs) is released from rest and subsequently slides (not rolls) through a curved arc of radius 3 ft before striking an identical ball B (also 2 lbs). Ball B then moves on to collide with ball C (likewise 2 lbs). Find the final velocities of each ball assuming the motion is purely translational (i.e. no rolling) and that the surface is frictionless. The coefficient of restitution for all collisions is e = 0.85. Rework the previous problem, except this time the balls roll instead of sliding. Don't forget to account for the rotational kinetic energy (the rotational inertia for a solid sphere is 2/5MR^2 where M is mass of the ball and R is its radius). Also, recall that the relationship between angular speed and translational speed v = omega R.

Explanation / Answer

2) mass of each ball is M = 2lbs

let the radius of each ball be R.

Let the initial height of the ball A be h.

Let assume that the zero level of potential energy at the horizontal surface.

When ball A reaches the horizontal surface from a height h it loses its potential energy PE and gains kinetic enery KE.

Let us suppose ball A is moving with velocity v and rolling with angular speed on the horizontal surface after rolling down the arc and before striking ball B.

So v = R

or = v/R

Now KE of the ball = 1/2Mv2 + 1/2(2/5MR2)2 = 1/2Mv2 + 1/2(2/5MR2)(v/R)2 = (7/10)Mv2

Applying conservation of energy we get

Mgh =  (7/10)Mv2

or v = [(10/7)gh] ------------- ()

Now the ball A collides with this speed with ball B. Let after collision speed of A and B be vA and vB respectively.

Applying conservation of linear momentum, we have

Mv = MvA + MvB

or v = vA + vB -------- (1)

coefficient of restitution is e = 0.85

vB - vA = 0.85v ----------- (2)

from equation (1) and (2) we get

vB = 0.925v

vA = 0.075v ---------- (a)

Now, the ball B strikes ball C with speed vB. Let us suppose after collision, speeds of B and C be v1 and v2 respectively

Applying conservation of linear momentum, we have

MvB = Mv1 + Mv2

vB = v1 + v2 ---------- (3)

coefficient of restitution is e = 0.85, So,

v2 - v1 = 0.85vB ---------- (4)

from equations (3) and (4), we have

v1 = 0.075vB = (0.925)0.075v = 0.0693v ---------- (b)

v2 = 0.925vB = (0.925)0.925v = 0.855v ---------- (c)

So the speeds of balls A, B and C are 0.075v, 0.0693v and 0.855v respectively given by equations (a), (b) and (c) respectively.

where v = [(10/7)gh] as we saw above....

P.S.- I'm not calculating the final numerical values as I am not sure what the unit to be used for speeds - ft/s or metre/s.

Please specify this and I will do this without delay...or you can calulate it by finding the value of calculating

v = [(10/7)gh].

v = [(10/7)(32ft/s2)(3ft)] = 11.71ft/s

using this vlaue of v we get

speed of ball A after collision = 0.075v = 0.878ft/s

speed of ball B after collision = 0.0693v = 0.811ft/s

speed of ball B after collision = 0.855v = 10ft/s

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....

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