# Homework Help: Relative velocities - Velocity of impact

1. Apr 2, 2013

### memoguy

1. The problem statement, all variables and given/known data
https://dl.dropbox.com/u/51681259/physics_diag.png [Broken]

Object C & P are moving at the same constant velocity to the right, in the configuration of the image. The friction between object C and the ground is 0. Suddenly the friction increases to μ=0.9 between the ground and object C. This makes object C decelerate to a stop over 4m. At what velocity does object P impact the right internal side of object C. There is no friction between C & P.

d = 1m (distance between p right side and c internal right side)
μ = 0.9
s = 4m (stopping distance)

2. Relevant equations
I can fine object c's initial velocity by combining a couple of formulas:
F_net=ma
F_friction=μmg
v^2=u^2+2as

I find the initial velocity of C & P as being 8.4m/s.

Then I can find the acceleration of C, relative to the ground, after the increase in fiction which turns out to be:
-8.82m/s/s

And that means that relative to C, P accelerates forward at 8.82m/s/s. <IS THAT CORRECT??>

Then I take p's relative acceleration and the initial velocity and put them through v^2=u^2+2as
and find:
v^2=8.4^2+(2*8.82*4)

v=9.39
So my answer is that P impacts C at 9.39 m/s. Is this even close to correct?

Last edited by a moderator: May 6, 2017
2. Apr 2, 2013

### Staff: Mentor

This looks fine.

Both C and P are never moving faster than 8.4m/s, and they are moving in the same direction. There is no way the impact can occur with more than 8.4m/s.

The calculation has two problems:
- the initial relative velocity between P and C is zero, not 8.4m/s
- the distance is just 1m and not 4m.

3. Apr 2, 2013

### memoguy

Sorry! That was so stupid!!

v^2=0^2+(2*8.82*1)
v^2 = 17.64
v=4.2 m/s

Is that correct?

4. Apr 3, 2013

### Staff: Mentor

That looks good.