# Kinetic energy of balls after collision

1. Nov 28, 2004

### h6872

If the first ball in problem 7) has a mass of 10 kg and the second ball is travelling 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, travelling 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

2. Nov 29, 2004

### HallsofIvy

Staff Emeritus
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