# Conservation of Momentum Problem?

1. Oct 14, 2009

### r_swayze

A cat stands on a skateboard that moves without friction along a level road at a constant velocity of 2.00 m/s. She is carrying a number of books. She wishes to stop, and does so by hurling a 1.20 kg book horizontally forward at a speed of 15.0 m/s with respect to the ground. (a) What is the total mass of the cat, the skateboard, and any remaining books? (b) What mass book must she now throw at 15.0 m/s with respect to the ground to move at −2.00 m/s?

I got the answer to part A, here is my work for part A:

m1vi1 + m2vi2 = m1vf1 + m2vf2

m1 = total mass of the cat, the skateboard, and any remaining books
vi1 = 2.0 m/s
vf1 = 0 m/s
m2 = 1.2 kg book
vi2 = 2.0 m/s
vf2 = 15.0 m/s

m1vi1 = -m2vi2 + m1vf1 + m2vf2

m1 = (-m2vi2 + m1vf1 + m2vf2) / vi1

m1 = (-1.2*2.0 + m1*0 + 1.2*15.0) / 2.0
m1 = 7.8 kg

Now for part B, I thought I had the correct answer but apparently it is wrong. Here is my work:

From part a
m1 + m2 = total mass of of the cat, the skateboard, and any remaining books
m1 + m2 = 7.8 kg + 1.2 kg
m1 + m2 = 9 kg = mt

mtvi = m1vf1 + m2vf2

9*2 = m1*-2.0 + m2*15.0
18 = -2m1 + 15m2

So m1 + m2 should equal to total mass = 9 kg

m1 + m2 = 9
m1 = 9 - m2

Plug in to equation

18 = -2(9 - m2) + 15m2
18 = -18 + 2m2 + 15m2
36 = 17m2
m2 = 2.11 kg

Have I erred somewhere? If you plug in 2.11 kg and 6.88 kg, it comes out roughly correct. Should I be using 9 kg in part B?

Any help is much appreciated

2. Oct 15, 2009

### ehild

Your solution must be correct but a rounding error. m2=2.1176 kg, should be rounded to 2.12 kg.

ehild

3. Oct 15, 2009

### r_swayze

I tried 2.12 as well, it was still incorrect. Is my method correct?

4. Oct 15, 2009

### ehild

I think, your solution is correct, but we may misunderstand the text. Maybe it asks the mass of book that the cat has to throw after it has stopped, to gain -2 m/s velocity. Try it...

But also the official solution can be wrong.

ehild