Provincial Exam: Simple momentum explosion question

[Senjai]

Homework Statement

A 1.0kg physics puck is at rest when a small explosion breaks it into three pieces. A 0.50 kg piece, goes north at 10 m/s, a .3 kg piece goes east at 20 m/s. What is the magnitude of the momentum of the third piece.

Homework Equations

$$p = mv$$
Momentum Before = Momentum After

The Attempt at a Solution

So i know that the momentum before is zero, so the sum of the momentum on the x-axis is zero, and the sum on the y-axis is zero.

Because the first two pieces are at right angles to each other, i calculated the net momentum, or their resultant, and stated that the momentum of the third piece should be equal and opposite to maintain the law of momentum conservation.

So i determined: the first piece had a momentum of 5 kg m/s north, the second, 6 kg m/s east.

Plugged the values into the pythagorean theorum and got 7.8 kg m/s..

Which is an answer on the sheet, but it's wrong. The correct answer is 3.3 kg m/s

Could anyone tell me where i went wrong??

Regards,
Senjai

 

[Senjai]

*bump* :(

 

[kerimek]

I would like to commend you for your attempt at solving this problem. However, your approach may have been incorrect.

Firstly, you mentioned that the momentum before the explosion is zero, which is correct. However, you cannot simply add the momenta of the two pieces to determine the momentum of the third piece. This is because momentum is a vector quantity and it has both magnitude and direction.

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system remains constant before and after an event. In this case, the event is the explosion.

So, the momentum before the explosion is equal to the momentum after the explosion. We can write this mathematically as:

m1v1 + m2v2 = m3v3

Where m1, v1 are the mass and velocity of the first piece, m2, v2 are the mass and velocity of the second piece, and m3, v3 are the mass and velocity of the third piece.

We know the values of m1, m2, v1 and v2 from the problem statement. So, we can rearrange the equation to solve for m3:

m3 = (m1v1 + m2v2)/v3

Substituting the values, we get:

m3 = (0.5 kg * 10 m/s + 0.3 kg * 20 m/s)/v3

m3 = (5 kg m/s + 6 kg m/s)/v3

m3 = 11 kg m/s/v3

Now, to find the velocity of the third piece, we can use the Pythagorean theorem to find the resultant velocity:

v3 = √(10^2 + 20^2)

v3 = √(100 + 400)

v3 = √500

v3 = 22.36 m/s

Finally, we can substitute this value back into our equation for m3 to get:

m3 = 11 kg m/s/22.36 m/s

m3 = 0.49 kg

Therefore, the mass of the third piece is 0.49 kg and its momentum is 3.3 kg m/s, which is the correct answer.

In conclusion, the key to solving this problem is to use the law of conservation of momentum and to consider the direction of each piece's momentum