Confusion about momentum question

[dark-ryder341]

Homework Statement

Hmm, well, this isn't an actual homework question. It's more of a general one, but it does apply to homework. What I want to know is, for momentum, how do you know when to use sine and cosine for vectors? Likeeee, let me show you an example of what I mean:

This is a depiction of the following: two hockey players approach each other at an angle with different speeds. mass(m1)=90kg, velocity(v1)=10 m/s; (m2)=100kg, (v2)=15 m/s. They collide and stick together. The question asks to find the final velocity of the two.

Homework Equations

pi = m1v1i + m2v2i
pf = m1v1f + m2v2f >>> (m1+m2)vf

The Attempt at a Solution

Now, here's the beginning of how it's solved, according to my teacher.

pi = (90+10)i + 100(15cos(theta)i(hat) + 15sin(theta)j(hat)) = (900+1300)i(hat) + 750j(hat)

I'm going to stop here, as this is what I don't understand. How did he figure out that he should use 15cos(theta) and 15sin(theta) rather than just 15 m/s? And how did he know that i(hat) was using cosine, rather than sine? If you understand what I mean...I'm just not sure where the cos and sin, came from, in other words, and how he knew what order it went in. There is another question in my book where it's switched - i(hat) is sine and j(hat) is cosine.

Thanks for any help!

 

[rl.bhat]

First of all select x and y axis. In the given problem v1 is along x-axis with i as the unit vector. j is the unit vector along y-axis.
pi = m1v1*i + m2v2*cosθ*i + m2v2*sinθ*J
If A and B are the two vectors with an angle θ between them, then the component of B along A is B*cosθ and component perpendicular to A is B*sinθ. Same principle is used in the above problem.

 

[kerimek]

Hello, it seems like you are struggling with the concept of using sine and cosine for vectors in momentum problems. First, it is important to understand that sine and cosine are trigonometric functions that relate angles to sides of a right triangle. In physics, we often use these functions to break down a vector into its components in order to solve problems involving multiple directions.

In the example you provided, the hockey players are approaching each other at an angle, meaning their velocities are not just in one direction. In order to accurately calculate the final velocity, we need to break down their initial velocities into components using sine and cosine. This allows us to add the components in the x and y directions separately, and then combine them to get the final velocity.

Now, how do we know which function to use for each component? It depends on the direction of the vector. In your example, the x-direction is represented by the i(hat) unit vector, and the y-direction is represented by the j(hat) unit vector. The i(hat) vector is pointing in the positive x-direction, while the j(hat) vector is pointing in the positive y-direction. So, for the x-component, we use cosine because it is adjacent to the angle and represents the x-direction. For the y-component, we use sine because it is opposite to the angle and represents the y-direction.

I hope this helps clarify the use of sine and cosine for vectors in momentum problems. Remember, it is important to break down vectors into components when dealing with multiple directions. Keep practicing and you will become more comfortable with using these trigonometric functions in physics problems.