How do I calculate the total momentum after collision?

In summary, car A was traveling at a steady speed of 32mph when it was hit from the left by car B. Both cars had the same mass, and it can be assumed that they can each be modeled as a particle. The two cars coalesce after the collision and travel in a straight line at a direction of 75◦ to the original direction of car A.
  • #1
diorific
19
0

Homework Statement



I'm having difficulties trying to figure this out, can you please help? Many thanks.

Car A is traveling at a steady speed of 15ms−1 (approximately 32mph)
in a straight line when it is hit from the left by car B. Each car has the
same mass, and it can be assumed that they can each be modeled as a
particle. The two cars coalesce after the collision and travel in a straight
line at a direction of 75◦ to the original direction of car A, as shown in the
diagram below, where X is the point of collision.

8557409930_494ca7e6da_m.jpg



The diagram is a plan view, and k is directed vertically upwards.

(a) The suggested alignments for the unit vectors are parallel and
perpendicular to the direction of the coalesced cars, as shown in the
diagram. Write down the momentum of each car before the collision



Homework Equations





The Attempt at a Solution



Using the notation P.before and P.after for the total linear momentum before and after and vB for the velocity of B, we have

PA.before= m(15cos75)i + m(15sin75)j
PB.before= m*vB(i+J)

The thing is because i don't have the angle for B I'm not sure if the above is correct.
 
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  • #2
Your expression for PB.before has the same magnitude for i and j. This assumes a specific angle (45 degrees from i). If you don't have the angle, you should write something more general than that. (Hint, why did you write cos75 for PA.before?)

Did you want help with the whole question or just help writing down PB.before? You didn't actually state what your supposed to do in this problem, so I can't help you with the latter.
 
  • #3
Decompose the problem into [itex]p_i[/itex] and [itex]p_j[/itex].
 
  • #4
DimReg said:
Your expression for PB.before has the same magnitude for i and j. This assumes a specific angle (45 degrees from i). If you don't have the angle, you should write something more general than that. (Hint, why did you write cos75 for PA.before?)

Did you want help with the whole question or just help writing down PB.before? You didn't actually state what your supposed to do in this problem, so I can't help you with the latter.

Well, the question continues as follows, but I only wanted help for part a.

(b) The police estimate that the coefficient of friction between the road
surface and coalesced cars was 0.9, and that the distance traveled by
the coalesced cars was 17m. From these estimates, estimate the speed
of the coalesced cars just after the collision, with justification of your
approach, and quoting a reference to any formula that you use. [5]
(c) Write down the momentum of the coalesced cars just after the
collision. [1]
(d) Estimate the speed of car B in mph, explaining which principle you
use.

Then PB.Before=m*vb ? without the vector components?

I'll try to solve the rest of the question, if I have any problems I'll come back :)
 
  • #5
So, does anyone know the linear momentum for car B??
 
  • #6
diorific said:
PB.before= m*vB(i+J)
That's wrong. As DrClaude suggested, you can put different scalars in the I and j directions, or you drop the i and j and treat vB as a vector.
Either way, you need an expression for the total momentum after collision.
 
  • #7
If I put PB.Before=m*vb
This is why I'm confused, since PA.before has vector components I don't know how to add it up to PB.before to get the momentum after collision.
 
  • #8
I was re-reading the problem and shouldn't you take
diorific said:
it is hit from the left by car B
as meaning that B is traveling at 90° to A?
 
  • #9
it could be, but the diagram is not showing 90 degrees.
But if there is no other way of solving this i might do it this way.
 
  • #10
diorific said:
it could be, but the diagram is not showing 90 degrees.
But if there is no other way of solving this i might do it this way.

I wan't sure if you had made the diagram, or if it was part of the problem. Then, no, do not assume 90°. In the first part of the problem, you do not have enough information to give a numerical answer for car B. You will have to express it in terms of an unknown.
 
  • #11
DrClaude said:
I wan't sure if you had made the diagram, or if it was part of the problem. Then, no, do not assume 90°. In the first part of the problem, you do not have enough information to give a numerical answer for car B. You will have to express it in terms of an unknown.

then this is correct PB.Before=m*vb

Now I've done part B and the initial velocity after collision is 17.326

So how do I find the total momentum after collision??
 
  • #12
diorific said:
So how do I find the total momentum after collision??

If you have the velocity, don't you have the momentum?
 
  • #13
Yes, all sorted :)
 
  • #14
diorific said:
then this is correct PB.Before=m*vb

Now I've done part B and the initial velocity after collision is 17.326

So how do I find the total momentum after collision??
You know the combined mass, speed and direction after collision.
Btw, you could determine the component of B's initial momentum in the j direction from the information in part a. But you do need the info from part b to get the i component.
 

Related to How do I calculate the total momentum after collision?

1. What is linear momentum?

Linear momentum, also known as momentum, is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and its velocity.

2. How is linear momentum calculated?

Linear momentum is calculated by multiplying an object's mass (m) by its velocity (v). The formula for linear momentum is p = mv.

3. What is the law of conservation of linear momentum?

The law of conservation of linear momentum states that the total momentum of a system of objects remains constant unless acted upon by an external force. This means that the total momentum before an interaction is equal to the total momentum after the interaction.

4. How is linear momentum different from angular momentum?

Linear momentum is a property of an object's motion in a straight line, while angular momentum is a property of an object's rotational motion. Linear momentum is a vector quantity, meaning it has both magnitude and direction, while angular momentum is a vector quantity that also takes into account the rotational axis.

5. What are some real-life applications of linear momentum?

Linear momentum has many applications in everyday life, such as in sports like baseball, where a player's swing creates a large amount of linear momentum to hit the ball. It is also important in transportation, as the momentum of a moving vehicle must be considered for safe driving. Additionally, understanding linear momentum is crucial in engineering and designing structures that can withstand forces and motion.

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