Collision problem

Conversation started Sep 30, 2018 at 9:53.

Nobody recognizeable

Sep 30, 2018 09:53

@sammygerbil is the picture understandable. I'm sorry for being late.

sammy gerbil

@Nobodyrecognizeable Yes I can read it. Which question are you asking about? What is your difficulty?

Nobody recognizeable

@sammygerbil I will ask the question on top.

sammy gerbil

ok question 8.50. Oblique collision. Where are you stuck?

Nobody recognizeable

@sammygerbil what does line of centers mean?

sammy gerbil

Draw the billiard balls in contact, with one centre higher up than the other. Draw a line connecting the centres of the two balls. The initial velocity of the incoming ball is not along this line, it is perhaps horizontal. You have to resolve its velocity into components along ang perpendicular to this line.

Nobody recognizeable

Sep 30, 2018 10:07

@sammygerbil which angle is 45 degrees here?

sammy gerbil

The angle between the velocity vector V1 and the line of collision.

Nobody recognizeable

@sammygerbil like this ?

sammy gerbil

No the blue arrow is just a pointer, the line of collision is the red line. The angle you want is between V1 and the red line.

Nobody recognizeable

Then along the line of collision component of velocity is$ v_1 cos 45$

@sammygerbil ^^

sammy gerbil

Yes that is correct. (Ignore V2 because in this case the 2nd ball is initially stationary I presume.)

So now the relative velocity of approach is $V_1\cos45$.

Nobody recognizeable

Sep 30, 2018 10:19

@sammygerbil ok .

@sammygerbil then what after that?

@sammygerbil why did you delete that ?

sammy gerbil

Sorry, I made a mistake there. I forgot $e \ne 1$. That comment only applied for $e=1$.

So the next step is to conserve momentum along the line of collision.

Nobody recognizeable

@sammygerbil ok fine go on man.

@sammygerbil how would what angle the balls will go so how can write the conservation of momentum along the line of action. Should I assume that angle $\theta $ or so ?

sammy gerbil

It is best to do the main part of the calculation as a head-on collision along the line of centres. Momentum conservation : $mU_1=mU_1'+mU_2'$ where $U_1=V_1\cos45$ is the initial velocity of ball 1 along the line of centres and $U_1'$ its final velocity. $U_2'$ is the final velocity of ball 2 in this direction - its initial velocity was zero.

Nobody recognizeable

@sammygerbil ok fine. What's next ?

sammy gerbil

So we've got $U_1=U_1'+U_2'$ from momentum conservation and relative velocity of separation = $U_2'-U_1'=eU_1$ from the Restitution Law.

From these 2 simultaneous equations you can find $U_1', U_2'$ in terms of $U_1=V_1\cos45$.

Nobody recognizeable

Sep 30, 2018 10:33

@sammygerbil in the second equation is $U_1 = vcos45 $ ?

sammy gerbil

In both equations $U_1=V_1\cos45$.

Actually you don't need to know $U_2'$ so there is no need to calculate that. Only $U_1'$.

Nobody recognizeable

@sammygerbil i got $ v_1' = \frac{v_1}{4\sqrt 2 }$

Whats next @sammygerbil

sammy gerbil

$V_1'$ here is the final velocity of ball 1 along the line of collision. This ball still has component of velocity $V_1\sin45$ perpendicular to the line of collision.

Nobody recognizeable

@sammygerbil i meant i got $U_1' = \frac{v_1}{4\sqrt 2 }$

sammy gerbil

Yes that is correct.

The ball still has velocity $\frac{V_1}{\sqrt2}$ perpendicular to the line of collision. So you need to add these to get the final velocity of ball 1.

You don't need to find the magnitude of final velocity, only its direction.

Then compare that with the initial direction (horizontal).

Nobody recognizeable

Sep 30, 2018 10:49

@sammygerbil ok then do I have to conserve the momentum along the initial velocity direction?

sammy gerbil

No. We have conserved momentum along the line of collision. Momentum is also conserved perpendicular to this direction. That means the velocity of ball 1 in this direction remains the same. This component is not affected by the collision.

Nobody recognizeable

@sammygerbil do i need to break the components of final velocity in horizontal?

sammy gerbil

Maybe. What I did was find the angle between the final velocity and the component perpendicular to the line of collision. This angle is $\theta$ where $\tan\theta=\frac14$.

Nobody recognizeable

@sammygerbil they've said tan$\theta = 3/5$

sammy gerbil

My angle $\theta$ is a different angle. The angle which is required is $45-\theta$.

So $\tan(45-\theta)=(\tan45-\tan\theta)/(1+\tan45\tan\theta)=3/5$.

Nobody recognizeable

Sep 30, 2018 11:02

@sammygerbil how do you get the $\theta$ please show that .

sammy gerbil

The final velocity component parallel to the line of collision (opposite) is $\frac{V_1}{4\sqrt2}$ as we calculated above. The perpendicular component (adjacent) is $\frac{V_1}{\sqrt2}$ which was not changed by the collision. The angle between these is $\theta$ where $\tan\theta=\text{opposite/adjacent}=\frac14$.

Nobody recognizeable

@sammygerbil i do get it from the triangle. But here comes a misconception which I have. Now do you have little more time for solving that ?

sammy gerbil

Nobody recognizeable

@sammygerbil as you said that you broke two orthogonal components. If they're orthogonal shouldn't they be perpendicular?

sammy gerbil

yes

Nobody recognizeable

Sep 30, 2018 11:13

@sammygerbil this angle should also be the angle between the vectors.

sammy gerbil

Which angle? Which vectors?

Wait, I think I see what you are thinking. The final velocity is the vector sum of the 2 components (parallel + perpendicular).

harambe

@JohnRennie@sammygerbil good morning

@JohnRennie are you free for some minutes

Nobody recognizeable

@sammygerbil yep.

sammy gerbil

@Nobodyrecognizeable The angle $\theta$ is between the final velocity and the perpendicular component.

And we know the perpendicular component is 45 degrees to the initial velocity.

Conversation ended Sep 30, 2018 at 11:17.

all rooms

Sep '1830

Problem Solving Strategies

General chat for high school physics. For MathJax see meta.sta...

context of the conversation
join this room
about this room

Participants

sammy gerbil 50%
Nobody recognizeable 45%
harambe 3%

all times are UTC