What are the components and magnitude

[Panix]

Homework Statement

A 2.0kg object's momentum at a certain time is 10kg*m/s 37degrees vertically upward from due west.

What are the components and magnitude of its velocity vector at this time (in a frame in standard orientation)?

Homework Equations

P=mv
imp= dP

The Attempt at a Solution

I have no clue on setting this problem up..my professor does not show us example problems or anything, just goes over equations.

 

[IBY]

Look carefully at the momentum equation. See the term v there? Why not solve for v? That would give you the velocity component.

 

[tomcue]

As a scientist, it is important to have a solid understanding of the equations and principles involved in solving a problem. In this case, the problem involves calculating the components and magnitude of an object's velocity vector at a certain time, given its mass and momentum. To begin, let us first define the variables involved in this problem:

m = mass of the object (2.0kg)
v = velocity of the object
P = momentum of the object (10kg*m/s)
θ = angle of the velocity vector (37 degrees)

The momentum of an object can be calculated using the equation P = mv, where m is the mass and v is the velocity. In this case, we know that the momentum is 10kg*m/s, so we can rearrange the equation to solve for v:

v = P/m = (10kg*m/s) / (2.0kg) = 5m/s

Now that we have the magnitude of the velocity, we can use basic trigonometry to calculate the horizontal and vertical components of the velocity vector. Since the velocity vector is 37 degrees vertically upward from due west, we can use the following equations:

vx = v*cosθ = (5m/s)*cos(37 degrees) = 4m/s
vy = v*sinθ = (5m/s)*sin(37 degrees) = 3m/s

Therefore, the components of the velocity vector are vx = 4m/s and vy = 3m/s. The magnitude of the velocity can be calculated using the Pythagorean theorem:

|v| = √(vx^2 + vy^2) = √(4^2 + 3^2) = √25 = 5m/s

Therefore, the components of the velocity vector are vx = 4m/s and vy = 3m/s, and the magnitude of the velocity vector is 5m/s. In terms of standard orientation, the object is moving 4m/s horizontally and 3m/s vertically upward from due west. It is important to note that these calculations are based on the given information and may vary depending on the specific problem and its conditions. As a scientist, it is important to carefully analyze and understand the equations and principles involved in a problem in order to accurately solve it.