# Change in Momentum Problem

1. Apr 28, 2013

### Legerity

1. The problem statement, all variables and given/known data

In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the surface at 37000 km/h. The original speed of the comet at that time was about 40000 km/h, and its mass was estimated to be in the range (0.10-2.5) x 10^14 kg. Use the smallest value of the estimated mass. What change in the comet's velocity did this collision produce?

2. Relevant equations

ΔP

3. The attempt at a solution

I tried using P_initial = P_final and come out with the change in velocity of 0, or 80000 km/h, and they are both incorrect. Could someone help me as to where I am going wrong? I did:

(m_p)(v_p) + (m_c)(v_c) = V(m_p + m_c) and solved for V.

2. Apr 28, 2013

### cepheid

Staff Emeritus
As you can imagine, crashing the probe into something a TRILLION times its mass is going to have a very very small effect. I get 0 as well, if I use a calculator that doesn't have enough precision ;).

Also, are you meant to assume that this is a "head on" collision, i.e. that vp and vc initially have directly opposite directions? The problem doesn't say. EDIT: but that is what I assumed.

3. Apr 28, 2013

### Legerity

I guessed that my calc was giving me 0 because of that too!!! And about the opposite directions, I am not sure since it is not stated in the original problem at all. I did solve the problem though. I did (372 x 37000)/(10^13) = 1.38 x 10^-6 and that seemed to work. I do not know why though...

4. Apr 28, 2013

### cepheid

Staff Emeritus
That's wrong. Completely. You were right originally:

ptotal = mpvp + mcvc

= (372 kg)(37000 km/h) + (1e13 kg)(-40000 km/h)

vfinal = ptotal/m, where m = (372 kg + 1.0e13 kg)

This should give you a final speed that is just a tiny bit less then 40000 km/h, in the negative direction. So the difference in speed of the rock will be some tiny number of order 10-6, but not what you got above.

5. Apr 28, 2013

### Legerity

The 1.38 x 10^-6 answer somehow is right because I inputted it into my online homework, and it said it was the correct answer! And using your method, my calculator always spits out 40000 km/h for the velocity, and I guess it is rounding a lot for that to happen?

6. Apr 28, 2013

### haruspex

If you note straight away that the probe's mass is so small compared with that of the comet that we can ignore its contribution to the coalesced mass, two benefits flow. First, we no longer care whether the two given velocities are collinear:
mc Δv = mc(v - vc) ≈ mp vp
Second, it avoids taking a small difference between two large numbers, and the consequent loss of precision.

7. Dec 3, 2014

### Eng.m

Why you didn't write the answer in this web ,, al most of the people who know the answer just like to write a lot ,, you should know that in physic we learn by seeing the equations not by reading a lot of words

8. Mar 31, 2015

### KonaGorrila

The V_probe is relative to the comet so the comet has no V_i so the equation should be,
(P_probe - P_comet)/(mass_probe+mass_comet)= ΔV
since it is relative to the comet,
V_comet = 0 so P_probe/mass_total = ΔV

ΔV= 372*37000(km/h)/(372+1*10^13)=?

Last edited: Mar 31, 2015