To what angle does each ball rebound?

[viendong]

need help ASAP [solved] :)

I've been trying for hours... and couldn't get it right though

A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
Question: To what angle does each ball rebound.?
Answer: m =100 g >>> 79.3 degree
m = 200 g>>>>>14.7 degree
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1) find (y_0)A when the 100 ball is pulled to the left w/ 45 degree angle
(y_1)A = L(1-cos(theta))
=1(1-cos(45)
= 0.293
2) From that I find (V_1)A = sqr(2*g*(y_0)A)
(V_1)A = 2.40 m/s

3) Now I find the (V_2)A = (m1-m2)/(m1+m2)*(V_1)A
V_2)A = - 0.8 m/s

4) Now I find (Y_2)A = (V_2)A ^2/2*g

4) Now I need to find the angle that rebound for Ma=100 g

(theta) = arcosin(1-(Y_2)A /L) = 37 degree...

but the answer is 79.3 degree, please help me what I've done wrong?

 

[Andrew Mason]

A 100 g steel ball and a 200 g steel ball each hang from 1.00 long strings. After rest, the balls hand side by side, barely touching. The 100 g ball is pulled to the left until its string is at a 45 degree angle. The 200 g ball is pulled to a 45 degree angle on the right. The balls are released so as to collide at the very bottom of their swings
Question: To what angle does each ball rebound.?

Think of this in terms of an elastic collision between the two balls at the bottom. You can determine their speeds at the bottom using:
mgh = mv²/2

Since the collision speed is independent of m, v1 = -v2
Using conservation of momentum you can determine their respective velocities after collision. From that, determine their respective kinetic energies immediately after the collision. Using mgh = mv²/2 you can determine the maximum height the respective masses will reach.

AM

 

[wais]

To find the angle of rebound, you need to use the conservation of energy principle. The total energy of the system (balls) at the bottom of their swings should be equal to the total energy at the top of their swings. This means that the potential energy at the top (when pulled to a 45 degree angle) should be equal to the kinetic energy at the bottom (when they collide). So, using the formula for potential energy (PE = mgh) and kinetic energy (KE = 1/2mv^2), we can set up the equation:

PE(top) = KE(bottom)
mgh = 1/2mv^2

Since the masses are the same and the height (h) is the same for both balls, we can cancel those out. This leaves us with:

g = 1/2v^2

Now, we can solve for the velocity at the bottom (v) by plugging in the value for g (9.8 m/s^2). We get v = 4.43 m/s.

But, we need to find the angle of rebound, not the velocity. So, we can use the formula for velocity (v = ucos(theta)) to find the angle. Plugging in the values we know (v = 4.43 m/s and u = 2.40 m/s), we get:

4.43 m/s = 2.40 m/s * cos(theta)

Solving for theta, we get theta = 79.3 degrees. This is the angle of rebound for both balls.

I hope this helps! Let me know if you have any further questions.