Need help using algebra to solve physics equations

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The discussion focuses on solving a physics problem involving an elastic and completely inelastic collision between two balls of different masses. The user seeks help with algebraically finding the final velocities of the balls after the collision. They correctly apply the conservation of momentum and kinetic energy principles but struggle with the algebraic manipulation of the equations. A step-by-step breakdown is provided, illustrating how to simplify the equations and solve for the final velocities, ultimately confirming that the solutions are -0.4 m/s for the 5 kg ball and 1.6 m/s for the 7.5 kg ball. Understanding the algebraic process is emphasized as crucial for solving similar problems.
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My question is not so much what to do it is just i have problems solving for a certain equation.

A 5.00kg ball, moving to the right at a velocity of 2m/s on a frictionless table, collides head-on with a stationary 7.50kg ball. Find the final velocities of the balls if the collision is (a) elastic and (b) completely inelastic

for part A

I have m1v1f + m2v2f= m1vi1 +0
to solve for final velocity 1(the ball for 5kg) you use

vf1=(m1-m2/m1+m2)vi1

I really have no i do how to algebraically solve for vf1, can someone explain that to me because when i use the equation i get the right answer which is -.400m/s

To find vf2 we use the fact that it is an ellastic collision and the kinetic energy before and after is the same

1/2m1(vf1^2)+1/2m2(vf2^2)=1/2m1(vi1^2)+0
Using this i have to solve for vf2 but again i have a problem algebraically solving for vf2. But there is a similar example in the textbook so i used what they had to solve for vf1 and vf2 but they do not show the work

vf2=(2m1/m1+m2)vi1
answer should be 1.60m/s

My problem is not understand ellastic equations, it is how to solve the first equation for vf1 and the second for vf2. Can someone show me step by step how to use algebra to solve for vf2 with what i typed above,
 
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so i just did this in lab last week so I am able to help you...

1/2 m1(vi1^2) + 0 = 1/2m1(vf1^2) + 1/2 m2(vf2^2)

all the 1/2s cancel

m1(vi1^2) = m1(vf1^2) + m2(vf2^2)

momentum before = momentum after
m1(vi1) = m1(vf1) + m2(vf2) ---> vf1 = [m2(vf2) - m1(vi1)] / m1

plug vf1 solved above into equation

m1(vi1^2) = m1 ([m2vf2 - m1vi1] / m1)^2 + m2(vf2^2)

m1(vi1^2) = m1 [(m2vf2 - m1vi1)/m1][(m2vf2 - m1vi1)/m1] + m2(vf2^2)

m1(vi1^2)=m1/m2[m2^2(vf2^2) - 2m2vf2m1vi1 + m1^2(vi1^2)]+m2(vf2^2)

(m1/m2 = 1/m1 so multiply both sides by m1 to give the left side m1^2 and the right side m1m2(vf2^2) and everything else stays the same)

m1^2(vi1^2) =m2^2(vf2^2)-2m2vf2m1vi1 + m1^2(vi1^2) + m1m2(vf2^2)

the m1^2(vi1^2) on both sides cancel

0 = m2vf2(m2v2f - 2m1vi1 + m1vf2)

set m2v2f - 2m1vi1 + m1vf2 = 0
(please don't ask me why bc i didnt really understand bc my teacher just said well one of them must be equal to zero because the equation is equal to 0)

m2vf2 + m1vf2 = 2m1vi1

from here its easy to solve for vf2 then plug the solution into the momentum before = momentum after equation and solve for vf1

I had problems with the algebra for this too...i hate using variables and not numbers. I hope this helps and you can understand each step, its hard to type it all out. Good luck!
 
Last edited:
since yu have the masses sub for them in the 2 equations. you obtain
5 v1f + 7.5 v2f = 10 ...(1) and
(1/2) 5 v1f^2 + (1/2) 7.5 v2f^2 = (1/2) 20 ...(2)
now divide eq # (2) by (1/2) also
divide both eq by 5 to siplify - you get
v1f + 1.5 v2f = 2 ...(3) and
v1f^2 + 1.5 v2f^2 = 4 ...(4)
see how simple they became
solve # (3) for v1f and sub in # (4)
solve the quadratic you obtained from # (4)
you will get v2f
sub this value in # (3)
and here you go you obtain v1f
The answers I obtained for
v1f is - 0.4 m/s and for
v2f is 1.6 m/s
Best wishes.
 
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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