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Phy 231
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Do contact Admissions tomorrow and, if they can't help you successfully make the switch, email me.
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Phy 231
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Cons_Momentum_Question
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I cannot figure out the after collision velocity of the first ball after contact with the second ball from my conservation of momentum lab as follows. Here is the section including the notes.
-------->>>>>>>> velocity first ball before, first ball after, second ball after collision; mean +- std dev first ball before, after, 2d ball after
Your answer (start in the next line):
115.6 cm/s, 88.55 cm/s, 213.6 cm/s
116.5 cm/s, 114.6 cm/s
89.47 cm/s, 87.63 cm/s
216.01 cm/s, 211.25 cm/s
First row is the mean velocity of the first ball before collision, the mean velocity of the first ball after collision, and the mean velocity of the target ball after collision.
Second row is the velocity of first ball + std deviation of before collision, then the velocity of first ball - std devaiation before collision.
Third row is the velocity + std dev of the first ball after collision, then the mean velocity - std dev of the first ball after collision
Fourth row is the velocity after of the target ball with + std dev, followed by the - std deviation.
Velocities were calculated as follows for the mean calculations.
Vo1xave = 'dso1 / 'dt: = 22.81 cm/ 0.3948 s = 57.78 cm/s
Vave = (vo + vf) / 2, with vo = 0 cm/s
57.78 cm/s = (0 + vf) / 2
vfo1 = 2* 57.78 cm/s = 115.6 cm/s
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The final velocity is double the initial velocity if the initial velocity is zero and acceleration is uniformj.
For the horizontal motion of the ball as it falls to the floor, which is what you are analyzing here, the initial velocity is not zero. In fact the horizontal velocity changes by only a miniscule amount, and is to be regarded as constant.
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I do know that horizontal velocity of the first ball is 115.6 cm/s before collision using 'dsy and 'dt for the time to drop that distance. is it still 115.6 cm/s after collision? or am i supposed to derivie it from a momentum calculation?
'dsy = 76.38 cm, and the time to drop that distance is 0.3948 s.
I know the velocity of the target ball using this information and the fact that the target ball has no initial velocity.
VaveTarget = 'dsTarget / 'dt = 42.17 cm / 0.3948 s = 106.8 cm/s
Using this info i can find the final velocity of the target ball.
vaveTarget = (vf + vo) /2 , with vo = 0cm/s i solve for vf
vf target = 2*vaveTarget = 2 * 106.8 cm/s = 213.6 cm/s.
Here is where i get stuck, I know the first ball decelerates upon impact with the second ball because it travels a lesser distance than it did with no collision. Intuition tells me to compare these two distances and corresponding velocities.
'ds no impact = 22.81 cm
'ds impact = 17.48 cm
The difference between the two is 22.81 cm - 17.48 cm = 5.33 cm
So the first ball after impact travels on average 5.33 cm less than without impact.
I want to use the conservation of momentum theorem of
m1*v1 + m2*v2 = m1*v1' + m2*v2'
but get stuck in the equation.
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For each ball in each situation you know the horizontal distance it traveled after leaving the ramp, and you know how far it fell.
The setup, if done correctly, ensures that the balls all begin their descent with zero, or near-zero, vertical velocity. So you can figure out how long they took to fall.
This allows you to calculate the horizontal velocity for each case. In each case the horizontal velocity is considered constant, since the small amount of air resistance encountered will have an insignificant effect on the motion of the ball.
Having obtained the correct horizontal velocities (three of which are nonzero and one of which, namely the velocity of the target ball before collision, is zero) you will have all four velocities to substitute into the equation.
At that point you will be able to manipulate the equation to find the ratio of the masses (of course if you get stuck on this you can send me your best thinking, along with the equation and your best attempt to rearrange it, and I'll be glad to advise you further).
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Perhaps i am supposed to use KE of the system and calculate from there.
KEo = 1/2 * m1 * vo^2 + 1/2 * mvoT^2 = = (m1 + m2) * (1/2) * vo^2 + vo^2T
KEo = (1/2) * (m1 + m2) * v01^2
And compare it against KE of both balls after
(1/2) * (m1 + m2) * vo1^2 = (1/2) * m1 * vf1^2 + (1/2) * m2 * vft^2
(1/2) * (m1 + m2) * vo1^2 = (1/2) * (m1 * v1'2 + m2 * vft^2)
(m1 + m2) * vo1^2 = m1 * v1^2 + m2 * vf2^2
So this is where i am stuck, I know there is some deceleration from impact, but i do not know how to calculate it.
is v1' still = 116.5 cm/s post impact?
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