Solve Conservation of Momentum Problems | Physics Homework Help

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The discussion revolves around solving conservation of momentum problems in physics, particularly focusing on two scenarios involving a man on ice and a collision between two balls. The principle of conservation of momentum is highlighted, emphasizing that the momentum before an event must equal the momentum after. Participants engage in clarifying the correct application of formulas and units, with specific attention to the mass conversions from weight and the correct algebraic manipulation needed to solve for unknown velocities. Misunderstandings about the equations and units are addressed, leading to a clearer understanding of how to approach these physics problems. The conversation underscores the importance of accurately applying physics principles and the need for careful algebraic work in problem-solving.
  • #31
OK, now try to do the same thing with your previous equation. It really is no different.

m_1V_1+m_2V_2=0

Can you show the following two steps?

1.) Subtract m_2V_2 from both sides.
2.) Divide by m_1.

Then you will have solved for V_1. But be careful about the negative sign that appears after step 1.
 
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  • #32
M1 + m2V2 = 0
M1 + M2V2 - M2V2 = 0 -M2V2
M1 / M1 = M2V2 / M1
 
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  • #33
Dr physics said:
M1 + m2V2 = 0

That is not the equation. Where did V1 go??
 
  • #34
o sry,

M1V1 + m2V2 = 0
M1V1 + M2V2 - M2V2 = 0 -M2V2
M1V1 / M1V1 = M2V2 / M1V1
 
  • #35
Dr physics said:
o sry,

M1V1 + m2V2 = 0

OK, that's the equation.

M1V1 + M2V2 - M2V2 = 0 -M2V2

Yes.

M1V1 / M1V1 = M2V2 / M1V1

No. First of all, what happened to that negative sign that I cautioned you to be careful about? You do understand that 0-M2V2=-M2V2, right? And second, why are you dividing by M1V1? You're supposed to be solving for V1. That's why I said in step 2, "divide by m_1"!
 
  • #36
im sorry i knew i shouldn't have, made a silly mistake sry about that.

M1V1 + m2V2 = 0
M1V1 + M2V2 - M2V2 = 0 -M2V2 = -M2V2
M1V1 / M1V1 = -M2V2 / M1
 
  • #37
Dr physics said:
im sorry i knew i shouldn't have, made a silly mistake sry about that.

M1V1 + m2V2 = 0
M1V1 + M2V2 - M2V2 = 0 -M2V2 = -M2V2

Right.

M1V1 / M1V1 = -M2V2 / M1

Again, you're dividing by M1V1. This is not right. The last line should read:

M1V1/M1=-M2V2/M1

Then we obtain V1=-M2V2/M1. You can then plug in the numbers you were given to find V1.
 
  • #38
M2 = 1.2kg
V2 = 5m/s
M1 = 750 N
V1 = ?

V1 = -M2V2/M1
V1 = -1.2 * -5 / 76.53 = 0.0784

is that correct?
 
  • #39
Dr physics said:
M2 = 1.2kg
V2 = 5m/s
M1 = 750 N

As I have already explained to you, M1 is not 750N. It is 76.53 kg. But you did use the correct number below, so this mistake is minor.

V1 = ?

V1 = -M2V2/M1
V1 = -1.2 * -5 / 76.53 = 0.0784

is that correct?

No, it isn't. The answer should obviously be negative. Think about what you're doing. If the book moves in one direction, then obviously the man has to move in the other direction! Why did you insert that extra negative sign in your last step? It shouldn't be there.
 
  • #40
V1 = ?

V1 = -m2v2/m1
V1 = -1.2 * 5 / 76.53
V1 = -6 / 76.53 = -0.0784
 
  • #41
Dr physics said:
heres a list of question I am having difficulty with.

1)define clearly the principle of conservation of momentum and apply this to solve the following problem:
A man weighing 730N stands in the middle of a frozen pound of a radius 5.9m. he is unable to get to the other side becuase of lack of friction between his shoes and the ice. to overcome this problem, he throws his 1.2kg book horizontally towards the north shore at a speed of 5m/s. how long does it take him to reach the south shore?

to work out how long it take him to reach the south shore do you use this formula

d = vi + vf / 2 * t = t = vi + vf / 2 / d

t = -0.0784 + 5 / 2 / 5.9
= 4.922 / 2 / 5.9
= 5.9 / 2.461
t = 2.34 s

hope that's right
 
Last edited:
  • #42
Dr physics said:
V1 = ?

V1 = -m2v2/m1
V1 = -1.2 * 5 / 76.53
V1 = -6 / 76.53 = -0.0784

Correct. And the units are m/s.
 
  • #43
Dr physics said:
to work out how long it take him to reach the south shore do you use this formula

d = vi + vf / 2 * t = t = vi + vf / 2 / d

No, you don't use that formula. In that formula, vi and vf refer to the initial and final velocities of the same object under uniform acceleration. You are trying to apply it to two different objects. You need to stop guessing and start thinking if you are going to succeed at this.
 

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