Solving Ellastic Collisions: Masses mA & mB, Velocity & Height

[Paulbird20]

Two balls, of masses mA = 40 g and mB = 72 g are suspended as shown in Figure 7-44. The lighter ball is pulled away to a 60° angle with the vertical and released.




(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)

(b) What is the velocity of each ball after the elastic collision?
ball A
ball B
(c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
ball A
ball B


Ok so i found part A by doing Square root of 2*9.8*.15 and got 1.71 which is correct.
Part B i got -.488 for the velocity of A and it is correct
Part B i got 1.22 for the velocity of B and it is correct

Where i am lost is how to find the heights of each.

I tried doing (.488^2)/ 2 *9.8*sin 60
and i get .01402 m
It is incorrect any help?

 

[saket]

Hint: How did u get "Ok so i found part A by doing Square root of 2*9.8*.15 and got 1.71 which is correct." ??

 

[Paulbird20]

sin 60 * 30 cm converted to meeters = the .15 and i think you can figure the rest

But i don't see where your going either

 

[saket]

sin 60 * 30 cm converted to meeters = the .15 and i think you can figure the rest

But i don't see where your going either

I was basically hinting at this "rest" only! Use the same approach.

Well, I guess, string length is given to be, l = 30cm. (I am not able to see any figure!)
In that case, height at an angle "theta", with the vertical, is l{1 - Cos(theta)}. Note that, Cos 60 = 0.5, Sin 60 = 0.866.

 

[Paulbird20]

that won't give me the right answer because i don't know the angle after colliision.

.3 (1-cos60)=.15 and that is not the height of either after collision

 

[saket]

sin 60 * 30 cm converted to meeters = the .15 and i think you can figure the rest

But i don't see where your going either

You are misinterpreting me. When I said, "Note that, Cos 60 = 0.5, Sin 60 = 0.866.", I was saying how could you possibly get "sin 60 * 30 cm converted to meeters = the .15". Instead, the correct formula should have been l{1 - Cos(theta)}.

Of course, you don't know the angle.. that can be calculated once height is known. To calculate height, use the same principle, which you used to obtain the result of part (A).. although in the reverse manner.

 

[Paulbird20]

ah you i was badly misunderstanding it. I got it now TY