Calculating Final Velocity of Mass 1 in an Elastic Collision

[ScienceGeek24]

Homework Statement

Mass m1=7kg with speed of 3 m/s collides at the origin with a mass m2= 8 kg that has a speed of 5 m/s, (at an angle of 33 degree in the 2nd qwuadrant). After the collision , mass 2 remains at rest at the origin. Find the final velocity of mass 1 (magnitude and direction).

Homework Equations

vf=m1+m2/m1-m1 time vx

vf=2m/m1+m2 vy

The Attempt at a Solution

I tried using this equation to get the right answer which is 3.59m/s by doind this (2(8kg)cos(33)/7kg+8kg time 5 m/s and i get 4.47 m/s. What I'm I doing wrong here?

 

[SammyS]

Homework Statement

Mass m1=7kg with speed of 3 m/s collides at the origin with a mass m2= 8 kg that has a speed of 5 m/s, (at an angle of 33 degree in the 2nd qwuadrant). After the collision , mass 2 remains at rest at the origin. Find the final velocity of mass 1 (magnitude and direction).

Homework Equations

vf=m1+m2/m1-m1 time vx

vf=2m/m1+m2 vy

The Attempt at a Solution

I tried using this equation to get the right answer which is 3.59m/s by doing this (2(8kg)cos(33)/7kg+8kg time 5 m/s and i get 4.47 m/s. What I'm I doing wrong here?

Be more specific about the direction of each object before the collision.

 

[ScienceGeek24]

Object 1 is at the orign of a coordinate system and object to is coming from the left to the right at an angle of 33 degree above the x axis.

 

[gneill]

The setup is something like this?

 

[ScienceGeek24]

yes but object 1 goes straight along the x axis.

 

[ScienceGeek24]

as a matter of fact m1 comes from the left side to the right at a speed of 3m/s. after the collision mass 2 remains at rest at the origin . Find the final velocity of mass 1 (magnitude and direction)

 

[gneill]

as a matter of fact m1 comes from the left side to the right at a speed of 3m/s. after the collision mass 2 remains at rest at the origin . Find the final velocity of mass 1 (magnitude and direction)

I think you should make your own drawing and post it, because your descriptions all seem to be self-contradictory to my (perhaps limited) understanding!

 

[ScienceGeek24]

ok this is the picture

 

[ScienceGeek24]

Mass m1=7 kg with a speed oof 3 m/s collides at the origin with a mass m2= 8 kg that has a speed of 5m/s, as shown. After the collision, mass 2 remains at rest at the origin. Find the final velocity of mass 1 (magnitude and direction).

 

[gneill]

Momentum must be conserved in the collision. Which body has nonzero momentum after the collision occurs? What are its component values?

 

[ScienceGeek24]

Object m1 one has non zero momentum after the collision. I would say the components are m1v1 I tried doign this. vf= 2(7kg)/ 7kg+8kg times 3m/s if is that what are you telling... it did not give me the answer.

 

[gneill]

Object m1 one has non zero momentum after the collision. I would say the components are m1v1 I tried doign this. vf= 2(7kg)/ 7kg+8kg times 3m/s if is that what are you telling... it did not give me the answer.

Momentum is a vector. It has components. What is the momentum vector of the system before the collision?

 

[ScienceGeek24]

I know that momentum is a product of mass times velocity and that velocity has direction. And the product of a scalar quantity and a vector quantity is a vector quantity. for m2 it would be M2cos(33)(5m/s) southwest, and m1 is m1cos(147)(3m/s) east. is that right?

 

[gneill]

The components of a vector should be parallel to the axes. That is, they'll be either positive or negative on the x-axis or y-axis.

Your drawing shows m1 initially traveling in the negative direction on the x-axis. So its only contribution will be along the x-axis:-m1*v1, where v1 is its speed (3m/s). Mass m2 is initially traveling more or less "southeast", and we'd expect it to have a positive contribution along the x-axis and a negative contribution along the y-axis. Use trig and the given angle to calculate its two components.

 

[ScienceGeek24]

ok so -(8kg)sin(33) + (8kg) cos(33) = 2.35 is that right??

 

[gneill]

ok so -(8kg)sin(33) + (8kg) cos(33) = 2.35 is that right??

The x and y components of the momentum must be kept separate. Also include the velocity;
Px = m*v*cos(θ), Py = m*v*sin(θ).

For these types of problem it is often best to start out by finding the x and y components of the velocities of the objects first. Then the momentum components can be obtained by multiplying each by their masses. Sum the like components of the momentum for each of the objects to find the total momentum components.

 

[ScienceGeek24]

Ok so you are saying that I have to sum Px = m2*v2*cos(θ) -m1*v1 then Py = m*v*sin(θ)-m1*v1?? sorry man i having trouble grasping what you are saying.

 

[ScienceGeek24]

i give up dude. I have no idea what you are saying. I don't see exactly what you mean. This one is too difficult.

 

[gneill]

You want to find the total momentum vector for the system. It will be a conserved quantity in any system.

Each individual body has its own momentum vector. You want to find those vectors and add them all together. To do this you first need the momentum vectors in component form (components in the x and y directions). Hence m*v*cos(θ) and m*v*sin(θ) for each one, using the appropriate values of m, v, and θ for each body.

Sum up all the x components. Sum up all the y components. That's your total momentum in its component form.

 

[ScienceGeek24]

I need to know if I'm doing anything right. I think that there is only one y component to sum up and that is just Py = m*v*sin(θ) and there is no other y component to sum up in the y direction other than that one, with that so there is only Py = m*v*sin(θ) for the y components. In the case of the x components i think i have to sum up Px = m2*v2*cos(θ)+-m1*v1 because they are both in the x direction. If I'm correct sum all up it would be m2*v2*cos(θ)-m1*v1+ m*v*sin(θ). If is THAT what you are telling me... I need to know if that is right. Or if i am doing anything right to what you are indicating me to look for...

 

[gneill]

I need to know if I'm doing anything right. I think that there is only one y component to sum up and that is just Py = m*v*sin(θ) and there is no other y component to sum up in the y direction other than that one, with that so there is only Py = m*v*sin(θ) for the y components.

That is correct. There is only the one nonzero component in this problem. Note that its direction points in the negative y direction, so it should be a negative value.

In the case of the x components i think i have to sum up Px = m2*v2*cos(θ)+-m1*v1 because they are both in the x direction.

Yes. That is correct.

If I'm correct sum all up it would be m2*v2*cos(θ)-m1*v1+ m*v*sin(θ). If is THAT what you are telling me... I need to know if that is right. Or if i am doing anything right to what you are indicating me to look for...

Sum the x and y components separately. Do not combine them. Momentum is a vector, so you should end up with separate x and y components for it.

Can you show your numerical values for the components of P?

 

[ScienceGeek24]

Py = m*v*sin(θ)= -(8kg)(5m/s)sin(33)
Px = m2*v2*cos(θ)+-m1*v1= (8kg)(5m/s)cos(33)-(7kg)(3m/s)

 

[ScienceGeek24]

is that right?

 

[gneill]

Py = -m2*v2*sin(θ)= -(8kg)(5m/s)sin(33)
Px = m2*v2*cos(θ)+-m1*v1= (8kg)(5m/s)cos(33)-(7kg)(3m/s)

That looks fine. What numbers will you put to Px and Py?

 

[ScienceGeek24]

I would put for Py=-21.78 kg/m/s

I would put for Px=12.54 kg/m/s

Note: what happened with the red number?? isn't that correct?? why would it be 1 instead of 2 if m1 has no y components?

 

[gneill]

I would put for Py=-21.78 kg/m/s

I would put for Px=12.54 kg/m/s

Yes. That's excellent.

Note: what happened with the red number?? isn't that correct?? why would it be 1 instead of 2 if m1 has no y components?

I added in the red bits because they appeared to be missing in your original.

Where will the momentum be after the collision (in which body)? Can you find its velocity vector?

 

[ScienceGeek24]

I think the momentum will be at object m1. How to find the velocity vector... hmm no i think i need help on that one.

 

[gneill]

I think the momentum will be at object m1. How to find the velocity vector... hmm no i think i need help on that one.

What's the formula for momentum?

 

[ScienceGeek24]

p=mv

 

[ScienceGeek24]

But this is an ellastic collision wouldn't it be vf=(2m/m1+m2)vx and vf=(m1-m2/m1+m2)vy ??

 

[gneill]

p=mv

So you have p, and you have m...

 

[ScienceGeek24]

hmmm still not 3.59m/s i divided the momentum of px/m=v and it gave me 1.79m/s ...

 

[gneill]

But this is an ellastic collision wouldn't it be vf=(2m/m1+m2)vx and vf=(m1-m2/m1+m2)vy ??

What would vx and vy be in that formula? Both objects are initially moving. That formula might apply if one object was moving and the other stationary.

Conservation of momentum always works.

 

[gneill]

hmmm still not 3.59m/s i divided the momentum of px/m=v and it gave me 1.79m/s ...

The velocity, like momentum, has components. Divide both of the momentum components by m1 to give the components of the velocity. Then, to find the speed, find the magnitude of that vector.

 

[ScienceGeek24]

Got it! Thanks man thankyou for your patience.

 

[gneill]

Got it! Thanks man thankyou for your patience.

No problem. Glad to help. Good luck!

 

[ScienceGeek24]

Wait! on emore thing! I'm trying to find the kinetic enrgy lost i this one and this is what i did KEf= 1/2(8kg+7kg)(3.52)^2=95.052 j and Ke total= 1/2(8kg)(5m/s)^2+1/2(7kg)(3m/s)^2=131.5 J than i subtracted Ktotal -KEf= 44... something something and that is not he right answer the right asnwer is 86.4J . What did i do wrong??

 

[gneill]

wait! On emore thing! I'm trying to find the kinetic enrgy lost i this one and this is what i did kef= 1/2(8kg+7kg)(3.52)^2=95.052 j and ke total= 1/2(8kg)(5m/s)^2+1/2(7kg)(3m/s)^2=131.5 j than i subtracted ktotal -kef= 44... Something something and that is not he right answer the right asnwer is 86.4j . What did i do wrong??

Only mass m1 is moving after the collision. Don't sum the masses, they are not stuck together.

KE1 = 131.5 j
KE2 = 45.1 j

KE1 - KE2 = 131 - 45.1 = 86.4 j

 

[ScienceGeek24]

which mas did you get the 45.1 with? because i did 1/2(7kg)(3.56)^2=44.35 not 45.1.

Sorry to take all your time man... :P

 

[ScienceGeek24]

opps never mind got it! Thanks gain man!