Kinetic energy of balls after collision

[h6872]

If the first ball in problem 7) has a mass of 10 kg and the second ball is traveling at it with a speed of 14 m/s, what is the velocity, to one decimal place, of the first ball after collision?

Problem (7) says A 9.8 kg steel ball, traveling left to right at 12 m/s, undergoes an elastic collision with a 5 kg steel ball at rest. To the nearest Joule, what is the Total Kinetic Energy of the two balls after collision?

So I derived equations using the conservation of momentum and kinetic energy and I got the two formulas:
from Ek = Ek1+ Ek2
1) 980 = 10 v1(prime)^2 + 5 v2 (prime) ^2
-------------- ---------------
2 2

from m1v1 + m2v2 = m1v1 (prime) + m2v2 (prime)
2) 190 = 10v1(prime) + 5 (v2) (prime)

so then I divided them all by 5 and rearranged to solve for v2(prime) getting:
v2 (prime) = 2v1(prime) - 38

I substituted this eqn back into the first and got
980 = 10v1(prime)^2 + 5 (2v1(prime) - 38)^2
This rearranges into a quadratic eqn (i think) the after I solve for v1 prime I keep getting the wrong answer! I'm not sure if it's just a math error or if I interpreted the question wrong... please help me!

thank you

 

[HallsofIvy]

I would interpret the (second) problem as saying that the first ball from problem 7 now has mass 10kg (instead of 9.8) but is still moving at 12 m/s while the second ball, of mass 5 kg, which was previously still, is now moving at 14 m/s.

That makes the total kinetic energy (1/2)(10)(12²)+ (1/2)(5)(14²)= 1210 Joules, not 980. The total momentum is (10)(12)- (5)(14)= 50 kg m/s

 

[wais]

for sharing your work and thought process! It seems like you have set up the equations correctly using the conservation of momentum and kinetic energy. However, it is possible that there may be a math error in your calculations. I would suggest double checking your calculations and also checking your units to make sure they are consistent throughout the equations.

As for the specific question about the velocity of the first ball after collision, we can use the same equations but with the given values of mass and initial velocity. So, we have:

From Ek = Ek1 + Ek2:
980 = 10v1(prime)^2 + 5 (14)^2
980 = 10v1(prime)^2 + 980
10v1(prime)^2 = 0
v1(prime) = 0 m/s

This means that the first ball will come to a complete stop after the collision with the second ball. This result makes sense since the second ball is initially at rest and the collision is elastic, meaning there is no loss of kinetic energy.

I hope this helps and good luck with your calculations!