Conservation of Momentum Problem?

AI Thread Summary
A cat on a skateboard moving at 2.00 m/s throws a 1.20 kg book at 15.0 m/s to stop, leading to a calculated total mass of 7.8 kg for the cat, skateboard, and remaining books. In part B, the user attempts to find the mass of a book needed to achieve a velocity of -2.00 m/s after stopping, resulting in a calculated mass of 2.11 kg, which rounds to 2.12 kg. However, there is uncertainty about whether the mass refers to a book thrown after stopping or during motion. The discussion highlights potential misunderstandings in the problem's wording and the possibility of an error in the official solution. Clarification on the problem's requirements is necessary for accurate resolution.
r_swayze
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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 into 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
 
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Your solution must be correct but a rounding error. m2=2.1176 kg, should be rounded to 2.12 kg.

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

ehild

I tried 2.12 as well, it was still incorrect. Is my method correct?
 
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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