Conservation of Momentum Problem?

In summary, the cat throws a 1.20 kg book horizontally forward at a speed of 15.0 m/s with respect to the ground to move at −2.00 m/s.
  • #1
r_swayze
66
0
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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  • #2
Your solution must be correct but a rounding error. m2=2.1176 kg, should be rounded to 2.12 kg.

ehild
 
  • #3
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?
 
  • #4
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
 

1. What is the principle of conservation of momentum?

The principle of conservation of momentum states that the total momentum of a closed system remains constant. This means that in the absence of external forces, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

2. How is momentum defined?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

3. How is conservation of momentum used to solve problems?

Conservation of momentum can be used to solve problems involving collisions or interactions between objects. By applying the principle of conservation of momentum, the initial and final momenta of the objects can be equated, allowing for the calculation of unknown velocities or masses.

4. Why is conservation of momentum important in science?

Conservation of momentum is a fundamental principle in physics and is important because it allows us to predict the outcomes of interactions between objects. It also helps us to understand the behavior of objects in motion and the effects of external forces.

5. Are there any real-life examples of conservation of momentum?

Yes, there are many real-life examples of conservation of momentum. Some examples include the recoil of a gun after firing a bullet, the motion of billiard balls colliding on a pool table, and the movement of rockets in space. Conservation of momentum is also important in sports, such as in a game of baseball when a pitcher throws a ball to a batter.

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