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phy 201
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 **
Gravitation
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PE = - G M m / r
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You have given me this equation previously but what is the G is it the gravity of earth or the universal gravitational constant?
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Small g, often italicized, stands for the acceleration of gravity at the surface of the Earth.
Large G stands for the universal gravitational constant 6.67 * 10^-11 N m^2 / kg^s.
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#$&*
phy 201
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 **
Momentum
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The masses of both balls are unknown. Using momentum conservation, you will determine the ratio of their masses:
Let m1 stand for the mass of the large ball and m2 the mass of the small ball. In terms of m1 and m2 write expressions for each of the following:
The momentum of the first ball immediately before collision, using the velocity you reported above (the velocity based on the mean range and distance of fall). Be sure to use both the numerical value of the velocity and its units. This will be reported in the first line below.
The momentum of the first ball immediately after collision, using the velocity you reported above. This will be reported in the second line below.
The momentum of the second ball immediately after collision, using the velocity you reported above. This will be reported in the third line below.
The total momentum of the two balls immediately before collision. This will be reported in the fourth line below.
The total momentum of the two balls immediately after collision. This will be reported in the fifth line below.
The total momentum immediately before collision is equal to the total momentum immediately after collision. Set the two expressions equal to obtain an equation. Report this equation in the sixth line below.
-------->>>>>>>> equation for momentum conservation
Your answer (start in the next line):
p = m1 * 35.0877 cm/s
p = m1 * 84.57 cm/s
p = m2 * 127.54 cm/s
Momentum before is just the momentum of the first ball
(m1 * 84.57) + (m2 *127.54) = p
(m1 * 84.57) + (m2 *127.54) = m1 * 35.0877
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Rearrange your equation so that all terms containing m1 are on the left-hand side, and all terms containing m2 are on the right-hand side. Report this equation in line 1 below.
Divide both sides by the appropriate quantity so that m1 appears by itself on the left-hand side. Report the resulting equation in line 2.
Divide both sides of the equation by m2, and report the resulting equation in line 3.
Simplify the right-hand side, if you have not already done so, to obtain a single number. If you have done your calculation correctly, the units will cancel out. Report the resulting equation in line 4. The left-hand side will be m1 / m2 and the right-hand side will be a single decimal number or, if you prefer, a reduced fraction.
Starting in the fifth line discuss the meaning of the ratio m1 / m2.
-------->>>>>>>> equation solution in steps, meaning of ratio m1 / m2
Your answer (start in the next line):
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I dont get how we are supposed to end up with m1/m2 Ive tried solvin git out but i always end up stuck on where to go
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If the equation was, for instance,
5 m_1 + 3 m_2 = 7 m_1
you would rearrange to get
-2 m_1 = -3 m_2.
Dividing both sides by m_2 you would get
-2 m_1 / m_2 = -3.
Dividing both sides by -2 you would get
m_1 / m_2 = 3/2.
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