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ljs1215 · Nov 18, 2009, 10:01 AM

I'm :confused:I would like to know if I am setting up this equation correctly.

Q - What is the momentum of an automobile weight 9800 N when it is moving at 35 m/s to the south.

This is what I have done -- am I on the right track.

mg - 9800N
vi = 35 m/s
vf = ?

I'm looking for the final velociity?

my equation:

m = w/g = 9800x1000=9.80 E6/9.8 m/s = 1.00E6
= 1.00E6 x 35m/s = 3.5E7 kg m/s2

Will my answer be 3.5E7 kg m/s or is there another part that I have missed.

LJS

ebaines · Nov 18, 2009, 10:43 AM

Why do you multiply the 9800 N by 1000? That puts the weight of the car in grams, but you want it on Kilograms. So:

$ mass = \frac {weight} g = \frac {9800 N} {9.8m/s^2} = 1000 Kg $

Then momentum is calculated from m*v , where m is in Kg and v is in meters/sec squared. **EDIT - the correct units for **EDIT - the correct units for v is meters per second - Doh! ** is meters per second - Doh! **

Some advice: AWAYS include the units in your calculations. If you had done that with this problem you would have seen that your calculation is off by a factor of 1000.

ljs1215 · Nov 18, 2009, 11:18 AM

Originally Posted by ebaines

Why do you multiply the 9800 N by 1000? That puts the weight of the car in grams, but you want it on Kilograms. So:

$ mass = \frac {weight} g = \frac {9800 N} {9.8m/s^2} = 1000 Kg $

Then momentum is calculated from m*v , where m is in Kg and v is in meters/sec squared.

Some advice: AWAYS include the units in your calculations. If you had done that with this problem you would have seen that your calculation is off by a factor of 1000.

So will this be my answer

1000kg x 35 m/s = 35000 kg m/s

or
will this by my answer

1000kg x 35 squared = 187.08

ljs1215 · Nov 18, 2009, 11:29 AM

Just to clarify,

my final step is:

1000kg x 35 m/s = 35000
or
1000kg x 35 squared = 187.08

ebaines · Nov 18, 2009, 12:27 PM

Momentum is mass times velocity, and is in units of kg-meters per second, so no squaring needed. I apologize for an error in my earlier posting where I said that v is meters per second squared - that's obviosuly incorrect, as velocity is in meters per second. Sorry for the confusion.

ljs1215 · Nov 19, 2009, 04:58 PM

Thank you so much for you help, however I have reached another stumbling block. Can you please explain this problem to me?

Problem:

The positions of three particles, written as (x,y) coordinates, are: particle 1 (mass 4.0kg) at (4.0m 0m); particle 2 (mass 6.0kg) at (2.0m, 4.0m); particle 3 (mass 3.0kg) at (-1.0m, - -2.0m). What is the location of the CM.

Q-1 Does CM mean center mass?

Unknown008 · Nov 19, 2009, 07:53 PM

I would have said yes.

To solve this, I would have used the principle of moments. I would look for a pivot point between two points. Then with the help of a sketch, I would draw a line joining the 'pivot point' to the opposite vertex of the triangle formed by the masses. There are three such lines and where they meet is the centre of mass. Well, I hope that is the way to do it :o

ebaines · Nov 20, 2009, 10:43 AM

Unknown008 - your technique would work only if the 3 particles are all of equal mass.

To determine the center of mass you can break the problem into x and y axes. The x component of the CM is calculated by multiplying each particle's mass by the value of its x-coordinate, add these terms up, then divide by the total mass:

$ \bar x = \frac { \sum (m_i X_i)} {\sum m_i} $

Repeat for the y coordinate:

$ \bar y = \frac { \sum (m_i Y_i)} {\sum m_i} $

Unknown008 · Nov 21, 2009, 01:26 AM

Well, or... find the centre of mass of two of the points, then find the centre of mass of the combined centre of masses and the last point. Both seem correct to me. Ok, let me try...

Hey, I got the same answer as with using your method. Well, except that it's more precise. But my method works :D

I'm using moments, not using the midpoint between two vertices.