Calculating change in momentum

[hmvince]

Homework Statement

A mass(1) of 10000kg with a velocity 30m/s west collides with another mass(2) of 1200kg with a velocity 20m/s north. After the collision the two masses stick together. Determine the change in momentum of mass(2) before and after the collision.

Homework Equations

General momentum equations:
p = mv
Δp = p_f - p_i

The Attempt at a Solution

p_i (mass 2) = mv = 1200 * 20 = 24000sN North
final velocity was calculated to be 26.87m/s 274.6°T (using LCM and some trig)
Therefore, p_f (mass 2) = 1200 * 26.87 = 32244sN 274.6°T

to calculate change in momentum:
Δp = p_f - p_i
Δp = (32244sN 274.6°T) + (24000sN South)^{Note, changed to from North to south as it was -p_i}

Using a vector diagram, I found the resultant vector, Δp, to be 38621sN 226.7°T.I am not sure that I am doing this right though. After some searching on the internet, some people calculated change in momentum by adding initial and final vectors. But isn't that just net momentum?

please help!

 

[gneill]

Homework Statement

A mass(1) of 10000kg with a velocity 30m/s west collides with another mass(2) of 1200kg with a velocity 20m/s north. After the collision the two masses stick together. Determine the change in momentum of mass(2) before and after the collision.

Homework Equations

General momentum equations:
p = mv
Δp = p_f - p_i

The Attempt at a Solution

p_i (mass 2) = mv = 1200 * 20 = 24000sN North
final velocity was calculated to be 26.87m/s 274.6°T (using LCM and some trig)
Therefore, p_f (mass 2) = 1200 * 26.87 = 32244sN 274.6°T

The magnitudes of the values look fine. What does the "T" syntax mean in your angle specifications?

to calculate change in momentum:
Δp = p_f - p_i
Δp = (32244sN 274.6°T) + (24000sN South)^{Note, changed to from North to south as it was -p_i}

Using a vector diagram, I found the resultant vector, Δp, to be 38621sN 226.7°T.


I am not sure that I am doing this right though. After some searching on the internet, some people calculated change in momentum by adding initial and final vectors. But isn't that just net momentum?

A change (Δ) is calculated by taking the difference between final and starting values. So a vector difference (subtraction) should be involved. Note that this can be accomplished by changing the sign of the appropriate vector and then adding.

 

[hmvince]

the T means true, so 90°T would mean East.

and yeah, I did the sign changing thing so I could simply add the vectors (north to south)

thanks!

 

[gneill]

the T means true, so 90°T would mean East.

and yeah, I did the sign changing thing so I could simply add the vectors (north to south)

thanks!

Okay, I understand. I think that you may have some issues with the angles. You should be able to tell just from the initial conditions that the direction of motion after the collision will be in the 2nd quadrant, so the True angle associated with the velocity and momentum should be between 90 and 180 degrees.

https://www.physicsforums.com/attachment.php?attachmentid=45772&stc=1&d=1333287551

 

[hmvince]

The diagram you sent looks pretty good. But i think i am still correct. °T is measured clockwise from north, so the final angle of the blue and red mass' are at an angle of about 300°T (loooking at your diagram). thanks for your help!