A 10.0 kg ball rests suspended by the 5.0 m long weightless string. It experienc

Question

A 10.0 kg ball rests suspended by the 5.0 m long weightless string. It experiences a central perfectly elastic collision with the 10 gram ball traveling at 800m/s as shown in the picture above.
Calculate: 1. The velocity of both balls right after the collision; 2. The maximum change in height the big ball will gain after the collision; 3. The maximum deflection angle α for the large ball. Calculate: 1. The velocity of both balls right after the collision; 2. The maximum change in height the big ball will gain after the collision; 3. The maximum deflection angle α for the large ball.

Explanation / Answer

1. The velocity of both balls right after the collision;

m1: .010 kg, v1(i): 800 m/s
m2: 10.0 kg, v2(i): 0.00 m/s

v1(f)= [10.0kg-.010kg)/(10.0kg+.010kg)(0m/s)]+
[2(.010kg)/(10.0kg+.010kg)(800m/s)
The first half equals zero, so just solve for the second half.
v1(f)= 1.60 m/s

v2(f)= [2(10.0kg)/(10.01kg/0m/s)]+
[(.010kg-10.0kg)/(10.01kg)(800m/s)]
And the same with this one, the first half equals zero, so just solve for the second half.
v2(f)= -798 m/s
NOTE - This velocity is negative because the smaller ball is now moving in the opposite direction. Make sense?

And, finally, to make sure that you've gotten both final velocities correct, use the conservation of momentum.

(10.0kg)(0m/s) + (.010kg)(800m/s) = (10.0kg)(1.60m/s) + (.010kg)(-798m/s)
Essentially, 8 = 8, so that proves your calculated velocities are correct.
It's really something like 8 = 8.02, but that's because of rounding. If you use the velocities without rounding, 1.5984 m/s and -798.4016 m/s it will equal 8 on both sides.

Now for 2, the maximum change in height the big ball will gain after the collision;

You can use one of two equations, which are really the same.
1) mgh = 0.5mv^2
2) h = v^2/2g

For equation #1,
[(10.0kg)(9.8m/s^2)(h)] = (0.5)(10.0kg)(1.60m/s)^2
h will equal .131m

For equation #2, (if you just want to check to be sure you're correct),
h = (1.60m/s)^2 / 2(9.8m/s^2)
h will equal .131m, again.

And I'm still working on #3, but I'll be glad to explain it once I've finished.

Good luck
:)

And now I have the third part for you!

h = L - L (cos a) = L (1-cos a)

Your h = .131m, from the second part of the problem.
Your L = 5.00 m, which is given in the original problem.

.131m = 5.00m(1-cos a)
.0262 = 1 - cos a
-0.9738 = -cos a
cos^-1 (0.9738) = a
a = 1.513 degrees.

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