Momentum physics using vectors.

In summary: The initial momentum is simply the sum of the initial momenta of the two blocks:$$p_{initial}=\vec{p}_1+\vec{p}_2=m\vec{u}_1+m\vec{u}_2$$plugging in the numbers:$$p_{initial}=(0.5)(60\vec{\imath}+40\vec{\jmath})+(0.5)(-60\vec{\imath}-60\vec{\jmath})$$$$p_{initial}=30\vec{\imath}+20\vec{\jmath}-30\vec{\imath}-30\vec{\jmath}$$$$p_{initial}=0\vec{\imath
  • #1
logan 1234343
2
0

Homework Statement


Block 1 (0.5 kg) travels with an initial velocity of [ 60i ] + [ 40j ] m/s and then collides with block 2 (0.5 kg) traveling with an initial velocity of [ -60i ] + [ -60j ] m/s.
After the collision, block 1 has a final momentum of [ 0i ] + [ 5j ] kg*m/s. Assume that no external forces are present and therefore the momentum for the system of blocks is conserved.

What is the total initial momentum of the system of blocks?
i + j kg*m/s


What is the change in momentum for block 1?
i + j kg*m/s


What is the change in momentum for block 2?
i + j kg*m/s


What is the final momentum of block 2?
i + j kg*m/s
--------------------------------------------------------------------------------
What are the initial kinetic energies of block 1, block 2, and the system of blocks?
Block 1: J
Block 2: J
System of blocks: J


What are the final kinetic energies of block 1, block 2, and the system of blocks?
Block 1: J
Block 2: J
System of blocks: J


What is the change in kinetic energy of the system of blocks due to the collision?
J



Homework Equations



p=mv

The Attempt at a Solution



I tried P1x= -10-40*.5
p1y=0
p2x=2.5-40*.5
where i got the final velocity from m1v1 +m2v2 + etc. =m1v1 +m2v2 + etc.
 
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  • #2
It helps to be clearer about your notation:
If the initial momentum is ##\vec{p}## and the final is ##\vec{q}## then
##\vec{p}=\vec{p}_1+\vec{p}_2## right?

If ##\vec{u}## and ##\vec{v}## are the initial and final velocities respectively, then:
##\vec{p}_1=m_1\vec{u}_1## etc. but in your case ##m_1=m_2=0.5\text{kg}=m## so ##\vec{p}=m(\vec{u}_1+\vec{u}_2)##

Vector addition using unit vectors works like this:

if ##\vec{u}=u_x\vec{\imath}+u_y\vec{\jmath}## and ##\vec{v}=v_x\vec{\imath}+v_y\vec{\jmath}##;
then ##\vec{u}+\vec{v}=(u_x+v_x)\vec{\imath}+(u_y+v_y)\vec{\jmath}##

Rewrite your calculations like this and they'll probably make more sense.
 
Last edited:
  • #3
logan 1234343 said:

Homework Statement


Block 1 (0.5 kg) travels with an initial velocity of [ 60i ] + [ 40j ] m/s and then collides with block 2 (0.5 kg) traveling with an initial velocity of [ -60i ] + [ -60j ] m/s.
After the collision, block 1 has a final momentum of [ 0i ] + [ 5j ] kg*m/s. Assume that no external forces are present and therefore the momentum for the system of blocks is conserved.

What is the total initial momentum of the system of blocks?
i + j kg*m/s


What is the change in momentum for block 1?
i + j kg*m/s


What is the change in momentum for block 2?
i + j kg*m/s


What is the final momentum of block 2?
i + j kg*m/s
Are these your answers or just a "template"? Do you mean for each that the answer is supposed to be of the form __ i+ __ j kg*m/s where you are to fill in the blanks? If so, why have you not made any attempt to solve them yourself? Do you know what "momentum" is?

--------------------------------------------------------------------------------
What are the initial kinetic energies of block 1, block 2, and the system of blocks?
Block 1: J
Block 2: J
System of blocks: J


What are the final kinetic energies of block 1, block 2, and the system of blocks?
Block 1: J
Block 2: J
System of blocks: J


What is the change in kinetic energy of the system of blocks due to the collision?
J



Homework Equations



p=mv

The Attempt at a Solution



I tried P1x= -10-40*.5
p1y=0
p2x=2.5-40*.5
where i got the final velocity from m1v1 +m2v2 + etc. =m1v1 +m2v2 + etc.[/QUOTE]
So what did you get? Why are you not showing us what you did and what result you got?
 
  • #4
Yes this is a "template" for my class. I am to solve for each of the following. I have tried and simply do not understand. I missed this particular class due to illness and haven't managed to understand the material. Least not this part.
My answers weren't submitted because i figured they'd be irrelevant. The computer counted them as wrong. I showed how i got the answers.
 
  • #5
We don't understand your working as you have written it - we need to see how you are thinking about the problem more clearly in order to best help you. Did you follow the suggestions in post #2?
 

1. What is momentum in physics?

Momentum is a physical quantity that describes the amount of motion an object has. It is defined as the product of an object's mass and its velocity.

2. How is momentum calculated using vectors?

Momentum can be calculated using vectors by multiplying an object's mass by its velocity vector. This will result in a vector quantity representing the object's momentum.

3. What is the conservation of momentum?

The conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant. This means that the total momentum before a collision or interaction must equal the total momentum after the collision or interaction.

4. How does direction affect momentum in vector calculations?

In vector calculations, direction is an important factor in determining the overall momentum of a system. The direction of an object's velocity vector will determine the direction of its momentum vector. This means that even if two objects have the same speed, their momenta can be different if they are moving in different directions.

5. Can momentum be negative?

Yes, momentum can be negative. This occurs when an object is moving in the opposite direction to its positive direction, as defined by the chosen coordinate system. In vector calculations, negative momentum is represented by a vector in the opposite direction to the positive momentum vector.

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