Conservation of mechanical energy This is simple but I am puzzledNeed help

[beatenbob]

Hi everyone I am a newbie here. I have a question regarding the change in height when two ball collides. I have draw a picture and uploaded the attachment below.

Here are some Additional information:
Ball A rolls down a frictionless slope and collide with a stationary Ball B. Both balls move together up again the frictionless slope and achieved a height, h. Note that both balls have equal masses, what is the maximum height attained by the balls after the collision?

Well, I have tried and I get the height 0.5m as my answer. But, the answer in the book is 0.25m. Below is the solution taken from the book:
Initial PE = 2(Final PE)
mg(1) = 2(2m)gh
h = 0.25m

Whereas this is my teacher's solution:

Initial PE = 10m Joules

Velocity of Ball A just before it collides with Ball B:
10m = 1/2*m*v^2
v = 20^0.5
Then, the velocity is divided by 2 (I don't know why)
It becomes:
1/2(2m)(1/2*20^0.50)^2 = 2mgh
Then he get the answer h= 0.25m

Well,my solution is :
Initial PE(Ball A) = Final PE(Ball A and Ball B)
mgh = (2m)gh
h = 0.5m

**I just couldn't understand why the book wrote 2(2m), and so, my teacher, made the velocity halved during his calculation. Can somebody point me out and give me explanation why it is 2(2m) ??

Thanks much!

 

[Doc Al]

Well, I have tried and I get the height 0.5m as my answer. But, the answer in the book is 0.25m. Below is the solution taken from the book:
Initial PE = 2(Final PE)
mg(1) = 2(2m)gh
h = 0.25m

The book's answer is correct, but the solution isn't very helpful. The first step assumes that you know that energy is cut in half during the collision--but they really should prove that. (It's true though.)

Whereas this is my teacher's solution:

Initial PE = 10m Joules

Velocity of Ball A just before it collides with Ball B:
10m = 1/2*m*v^2
v = 20^0.5
Then, the velocity is divided by 2 (I don't know why)

Because during the collision momentum is conserved, not energy:
$$mv_i = (2m)v_f$$
$$v_f = v_i/2$$

It becomes:
1/2(2m)(1/2*20^0.50)^2 = 2mgh
Then he get the answer h= 0.25m

Right.

Well,my solution is :
Initial PE(Ball A) = Final PE(Ball A and Ball B)
mgh = (2m)gh
h = 0.5m

You assume conservation of energy, which doesn't apply during the collision. Momentum is conserved during the collision, but energy is lost.

From conservation of momentum we know that the speed is cut in half during the collision. Let's see what happens to the kinetic energy:
$${KE}_i = 1/2 m v_i^2$$
$${KE}_f = 1/2 (2m) v_f^2 = 1/2 (2m) (v_i/2)^2 = 1/2 {KE}_i$$

That's why the book can say that the initial PE is twice the final PE.

 

[beatenbob]

Okay thanks Doc Al...Your explanation makes sense to me..Thanks lot