2 balls involved in 1-D Elastic Collision

AI Thread Summary
In the discussion about a head-on elastic collision between two balls of different masses and velocities, participants explore the conservation of momentum and kinetic energy to determine their final velocities. The solution manual introduces the conservation of momentum equation and the relative velocity equation, which states that the difference in final velocities is equal to the initial relative velocity. Confusion arises regarding the interpretation of negative velocities, but it is clarified that the difference between two negative values can indeed be positive. The validity of the relative velocity equation is linked to Newton's Law of Restitution, which is applicable only in elastic collisions. The discussion emphasizes understanding these principles to solve collision problems effectively.
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A ball of mass 0.200 kg has a velocity of 1.50i m/s; a ball of mass 0.300 kg has a velocity of -0.400i m/s (where "i" is supposed to be that unit vector along x-axis). They meet in a head-on elastic collision. (a) Find their velocities after the collision.

I know I can use the conservation of both momentum and kinetic energy to solve this problem. I've done that. I looked in the solution manual and the person who wrote the solution used a different method I don't comprehend. First, they state the conservation of linear momentum, then they related the final velocities and way I don't understand.

The following 2 equations are used to solve the problem in the solution manual:

conservation of momentum for the two-ball system gives us:
0.200 kg (1.50 m/s) + 0.300 kg(-0.400 m/s) = 0.200 kg V1f + 0.300 kg V2f

Relative Velocity Equation:
V2f - V1f = 1.90 m/s


The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it? ( don't know, that's why I'm here, asking :-P ) So, how do they come to that conclusion?
 
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"The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it?"

Not necessarily: (-6)- (-7)= 1. The difference between two negative numbers can be positive.
 
SpeeDFX said:
Relative Velocity Equation:
V2f - V1f = 1.90 m/s


The reason this second equation confuses me is because, if both the balls had negative velocities after the collision, then V2f - V1f would be a negative velocity, wouldn't it? ( don't know, that's why I'm here, asking :-P ) So, how do they come to that conclusion?
Just because both balls have negative velocities does not mean their difference must also be negative. -3 and -4 are negative numbers but -3 - (-4) is not.

Nevertheless, there is a relevant question here : "Why is this second equation valid?" OR "Why must the relative velocity be reversed ?" Notice that this second equation is basically saying : v2 - v1 = u1 - u2 (assuming the first ball is the lighter one)

This second equation is often called an "empirical law" and goes by the name of Newton's Law of Restitution. It is an experimentally determined law and as yet, the best derivation I can think of (have not seen an elementary derivation of it) goes along the following lines :

Imagine you are an observer sitting on ball #2. We shall describe the dynamics of the collision relative to your frame of observation. Writing the energy conservation equation (in your frame, you are at rest and only ball #1 moves) ,we have :

KE~(before) = \frac{1}{2}m_1 ~u_{1-rel}^2~~~~~~~--~~~(1)

and

KE~(after) = \frac{1}{2}m_1 v_{1-rel}^2 ~~~~~~~--~~~(2)

where

u_{1-rel} = u_1 - u_2 ~~and~~v_{1-rel} = v_1 - v_2

Sunbstituting and equating (1) and (2) gives :

\frac {1}{2} m_1(u_1 - u_2)^2 = \frac {1}{2} m_1(v_1 - v_2)^2

(u_1 - u_2)^2 = (v_1 - v_2)^2

This allows two possibilities


u_1 - u_2 = v_1 - v_2
or
u_1 - u_2 = v_2 - v_1

Of these two, the second one must be correct as the first would lead to a direct contradiction of momentum conservation.

It is important to note that this form of Newton's Law of Restitution is valid only for an elastic collision. The more general form is :

\frac {v_1 - v_2}{u_1 - u_2} = e ~(coef.~of~restitution)~,~~-1 \leq e \leq 0

For a perfectly elastic collision, e = -1 and for a perfectly inelastic collision, e=0.
 
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Thanks a lot for the assistance you guys. this has abeen huge help :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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