# [Momentum and Energy] How much of initial velocity is lost?

1. Mar 20, 2017

### Notaphysicsmajor

1. The problem statement, all variables and given/known data
A 10kg block moving at 1 m/s hits another 10kg stationary block. Those two blocks (now stuck together) collide with another 10kg stationary block. What percent of the original energy is lost after the two collisions?

2. Relevant equations
m1v1 + m2v2 = (m1+m2)v3
Kinetic energy = .5(m)(v)^2

3. The attempt at a solution
Hello,

I'm not sure if I understand this problem correct so I hope someone here can help me, but here is how I solved it.

First collision:

(10)(1) + (10)(0) = (20)v
Velocity = .5 m/s

Second collision:

(20)(.5) + (10)(0) = (30)v
Velocity = .3

Initial energy (before any collision):

.5(10)(1)^2 = 5 J

Energy after first collision:

.5(20)(.5)^2 = 2.5 J

Energy after second collision:

.5(30)(.3)^2 = 1.6 J

1.6 J is amount of kinetic energy after both collisions, but problem asks what percent of the original energy was lost after first two collisions.

So 5 - 1.6 = 3.4 J energy lost after first two collisions.

3.4 / 5 = .68 or 68%. 68% of original energy was lost after first two collisions.

Is this correct?

2. Mar 20, 2017

### TomHart

Looks right to me, but I think I might have carried it out to one more decimal point. e.g. final velocity = 0.33 m/s

3. Mar 20, 2017

### PeroK

First, you need to be careful making numerical approximations. Sometimes you can use vulgar fractions to get an exact figure to work with. In this case: $\frac53 J$ is an exact number and is much better than $1.6 J$.

You can see that 68% is not quite right, because it ought to be $\frac23 \approx 66.7\%$

Second, what happens if you apply conservation of energy and momentum to the initial and final states and don't worry about calculating anything for the first collision?