Is the total magnitude of the momentum before the collision the same as the total magnitude of the momentum?

Question

Kathy and Bill collided at an intersection. The police learned the mass of the truck is 3200 kg and the mass of the car is 2800 kg. Based on the length of the skid marks at the scene and the mass of the vehicles, police estimate that the combined mass was moving at 7.0 m/s just after impact. From this point, the two vehicles slid and came to rest at the corner of the intersection.

Using the principle of conservation of momentum, the police determined that Bill’s truck was travelling at 2.3 m/s before the collision and Kathy’s car was travelling at 15 m/s before the collision.

The two momentum vectors before the collision are represented by the horizontal and vertical lines below. The momentum after the collision is the slanted line. Adding the two momentum vectors before the collision equals the momentum after the collision, according to the law of conservation of momentum

My answer is:

I don't understand how the momentum is different and what it means:

momentum before truck is 7293.22 kg m/s

momentum before car is 41361.93 kg m/s

Total momentum before is 48655.15 kg m/s

Total momentum after is 42000 kg m/s

Is the total magnitude of the momentum before the collision the same as the total magnitude of the momentum just after the collision? Explain why or why not.

Answer 1

The others have given the correct theoretical analysis. YOU should comment on what you have told us.

Is 48655.15 the same as 42000 ?

I can tell ( without sophisticated maths ) that it isn't.



So what you have deduced is that the arithmetic sum of the momenta prior to the collision is NOT the same as the momentum after the collision.



i.e "the total of the magnitudes of the momenta prior to the collision is NOT the same as the magnitude just after the collision"



The problem is in the misuse of the term "total"

The CORRECT question is " Is the magnitude of the momentum prior to the collision the same as the magnitude of the momentum after the collision"



Note that there is no such word as "total" in this question.

The momentum prior to the collision is a single number calculated by adding the momenta as vectors and finding the single equivalent vector then you may find the magnitude of this one.



If the two vehicles were at right angles at the point of collision then you may use Pythagoras' theory to find the magnitude of the (single) resultant vector.



C^2 = A ^ 2 + B^2

c = sqrt(7293.22^2 + 41361.93^2) = 42000.004 kg m/s



NOW is this the same as 42000? Yes or No.

Answer 2

momentum must be conserved... this is called the law of conservation of momentum..



here is a link to collisions... http://hyperphysics.phy-astr.gsu.edu/hbase/elacol.html



this shows that the difference between collisions is whether energy is conserved..



=====



the DIRECTION MATTERS.... I dont know the directions of the motions ... so the momentum of the car and the truck are at 90 degs to one another and the TOTAL momentum is vector addition...



combined mass is 3200 + 2800 = 6000 kg

and its moving at 7.0



so its momentum is 7.0*6000 = 42000



===========



truck before 2.3 m/s * 3200 = 7360



car before 15* 2800 = 42000



==================



assuming that they hit at 90 degs its hard to know from the description.. then the resultant vector will be some angle



sqrt ( 7360^2 + 42000^2) = 42640 this is the magnitude.. because the truck and the car were going different directions 90 degs to one another..



if you work in significant digits the momentums will probably be identical they are almost the same as is..



lets look at significant digits... 7.0 m/s is 2 significant digits..

3200 is 2 significant digits..



the answer will be 2 significant digits i think



http://www.physics.uoguelph.ca/tutorials/sig_fig/SIG_dig.htm



so the least significant digits is 2



so the answer can have only 2 significant digits for the combined mass and that is 42000 so that is correct..



=========



before it is 2 sign digits min so that is the number in the answer..



for the truck 7360 becomes 7.4 *10^2 because we round up the 3 because of the 6



and



for the car 42000 becomes 42000



so we get a bit more



sqrt(42000^2 + 7400^2) = 42646.



which in significant digits is 43000



so its a bit off but close...



I really dont know where you got your momentums of 42361.93 .. from the information given my numbers are correct..



and my total momentum is found because since the car and truck are moving at 90 degs the vectors they are components of the final vector



so the final vector mag is 42.6 *10^3 or 43 *10^3 which is a bit off from what you were given



I think the momentums are basically the same I think the error must be in the information given because your numbers are lower than mine so the momentums give a final of 42000



so good luck ... hope It helps..

Answer 3

"Is the total magnitude of the momentum before the collision the same as the total magnitude of the momentum just after the collision?"



NAY



Momentum is a vector... Magnitude AND direction



You have to break it down into horizontal/vertical components



Total initial horizontal momentum = Total final horizontal momentum

Total initial vertical momentum = total final vertical momentum



The Horizontal component and the vertical component make (2) legs of the triangle, so you cannot compare them to the hypotenuse like you are doing.



momentum before truck is 7293.22 kg m/s

momentum before car is 41361.93 kg m/s

Total momentum after is 42000 kg m/s



Assuming that the truck and car are initial perpendicular, use the Pythagorean Theorem to find the hyptonuse...

Total momentum before = sqrt(7293.22^2 + 41361.93^2) = 42000 kg m/s

Total momentum before = Total momentum after