- #1
beatenbob
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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!
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!
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