Motion of Particles: Showing u1 > (1+e)m2u2/(m1-em2)

[thereddevils]

Homework Statement

Particle A with mass m1 and particle B with mass m2 move with speed u1 and u2 respectively towards one another on a smooth horizontal plane . THe coefficient of restitution between particle A and particle B is e . If m1>em2 , and particle A continues to move without changing its direction after colliding with particle B, show that

u_1>\frac{(1+e)m_2u_2}{m_1-em_2}

Homework Equations

The Attempt at a Solution

some hints?

 

[kuruman]

Hint 1: What equation is always relevant when you have a collision?
Hint 2: What is the definition of the coefficient of restitution?

 

[thereddevils]

Hint 1: What equation is always relevant when you have a collision?
Hint 2: What is the definition of the coefficient of restitution?

thanks ,

m1u1+m2u2=m1v1+m2v2 since its an elastic collision .

From Newton's law of restitution ,

v2-u2=e(v1-u1)

do i play around with this two formulas ?

 

[kuruman]

thanks ,

m1u1+m2u2=m1v1+m2v2 since its an elastic collision .

Whether the collision is elastic or inelastic, momentum is conserved. In fact if the coefficient of restitution is anything but 1, the collision is inelastic, i.e. kinetic energy is not conserved.

From Newton's law of restitution ,
v2-u2=e(v1-u1)

Not quite. You need absolute values for the relative velocity.
|v₂-u₂| = e|v₁-u₁|

do i play around with this two formulas ?

Yes, but you have to use correctly the information that the two objects are moving in opposite directions before the collision but in the same direction after the collision. Be mindful of where you put plus and minus signs in front of each velocity. Note that the coefficient of restitution is essentially the ratio of relative speeds.

 

[thereddevils]

Whether the collision is elastic or inelastic, momentum is conserved. In fact if the coefficient of restitution is anything but 1, the collision is inelastic, i.e. kinetic energy is not conserved.

Not quite. You need absolute values for the relative velocity.
|v₂-u₂| = e|v₁-u₁|

Yes, but you have to use correctly the information that the two objects are moving in opposite directions before the collision but in the same direction after the collision. Be mindful of where you put plus and minus signs in front of each velocity. Note that the coefficient of restitution is essentially the ratio of relative speeds.

thanks for explaining,

so i think the first equation should be

m_1u_1-m_2u_2=m_1v_1+m_2v_2

I am not sure with this restitution part. What exactly is e ? epsilon?

I only know that Newton's law of restitution which states that

v_2-u_2=e(v_1-u_1)

If i include the modulus, i don find a way to solve the equations except if i square them, do I ?

 

[kuruman]

Your momentum conservation equation is correct with the understanding that v₁, v₂, u₁ and u₂ represent speeds, i.e. they are all positive quantities.

The coefficient of restitution is e, given by the equation

e=\frac{|v_2-u_2|}{|v_1-u_1|}

You don't need to square anything. As I said earlier, think of the numerator and denominator as the relative speeds. Speed is always positive.

In this case, before the collision the particles are moving in opposite directions, therefore the relative speed is the sum of the speeds of the two masses. That's a positive number. After the collision, the particles are moving in the same direction, which means that the relative speed is the difference of the speeds of each mass. That's also a positive number if you subtract the smaller speed from the larger one.

 

[thereddevils]

Your momentum conservation equation is correct with the understanding that v₁, v₂, u₁ and u₂ represent speeds, i.e. they are all positive quantities.

The coefficient of restitution is e, given by the equation

e=\frac{|v_2-u_2|}{|v_1-u_1|}

You don't need to square anything. As I said earlier, think of the numerator and denominator as the relative speeds. Speed is always positive.

In this case, before the collision the particles are moving in opposite directions, therefore the relative speed is the sum of the speeds of the two masses. That's a positive number. After the collision, the particles are moving in the same direction, which means that the relative speed is the difference of the speeds of each mass. That's also a positive number if you subtract the smaller speed from the larger one.

is the equation of restitution

e=\frac{V_2+U_2}{U_1-V_1} ?

 

[kuruman]

I am sorry, I misread the problem. I thought subscripts "1" and "2" referred to "before" and "after" the collision. Actually they label the particles. So if "u" stands for "before" and "v" stands for "after", the correct expression for the coefficient of restitution is

e=\frac{|v_2-v_1|}{|u_2-u_1|}

Start from there and, again, sorry about the confusion.

 

[thereddevils]

I am sorry, I misread the problem. I thought subscripts "1" and "2" referred to "before" and "after" the collision. Actually they label the particles. So if "u" stands for "before" and "v" stands for "after", the correct expression for the coefficient of restitution is

e=\frac{|v_2-v_1|}{|u_2-u_1|}

Start from there and, again, sorry about the confusion.

oh its ok , so the formula is now

e=\frac{v_1-v_2}{u_1+u_2}

I tried playing around with the equations but i don see a way to get rid of v_1 and v_2
since they are not required in the final answer .

 

[kuruman]

oh its ok , so the formula is now

e=\frac{v_1-v_2}{u_1+u_2}
I tried playing around with the equations but i don see a way to get rid of v_1 and v_2
since they are not required in the final answer .

Note that the problem asks you to show that u₁ is greater than a certain expression if the mass that has speed u₁ is to continue moving in the same direction after the collision.

Which way do you think the mass will move if it has speed less than that expression?
What about if it has speed equal to that expression?

Answer these two questions and you will see what to do with v₁ and v₂.

 

[thereddevils]

Note that the problem asks you to show that u₁ is greater than a certain expression if the mass that has speed u₁ is to continue moving in the same direction after the collision.

Which way do you think the mass will move if it has speed less than that expression?

i think the mass A will move to the left instead to the right with mass B moving to the left too


What about if it has speed equal to that expression?

Mass A will stop after collision and B will continue to move to the right ?

Answer these two questions and you will see what to do with v₁ and v₂.

But that tells me the direction of these particles , i still do not know their magnitudes .

 

[kuruman]

But that tells me the direction of these particles , i still do not know their magnitudes .

Think it through. If you can find what u₁ must be so that particle 1 is at rest after the collision, then any value of u₁ greater than that will give you the inequality that you want. Now, if particle 1 is at rest after the collision, you know v₁, do you not? So go ahead and find the threshold value of u₁.

 

[thereddevils]

Think it through. If you can find what u₁ must be so that particle 1 is at rest after the collision, then any value of u₁ greater than that will give you the inequality that you want. Now, if particle 1 is at rest after the collision, you know v₁, do you not? So go ahead and find the threshold value of u₁.

ok .

using the first formula , m1u1-m2u2=m1v1+m2v2

v1 is 0 ,

u_1=\frac{m_2v_2+m_2u_2}{m_1} ---1

then , from formula 2, put v1=0 too , v2+u2=eu1

put this into 1 , i got u1=(m2 e u2)/(m1)

nvm , this is definitely wrong but is this what you meant .

 

[kuruman]

ok .

using the first formula , m1u1-m2u2=m1v1+m2v2

v1 is 0 ,

u_1=\frac{m_2v_2+m_2u_2}{m_1} ---1

Correct.

then , from formula 2, put v1=0 too , v2+u2=eu1

Incorrect. Try solving for eu₁ again.

 

[thereddevils]

Correct.


Incorrect. Try solving for eu₁ again.

thanks Kuruman , i got it ... finally !