Ball and Box Collision: Finding Final Velocity

[Squizzel]

Homework Statement

A ball with mass 2.5 kg is moving in outer space with a velocity of 6.0 m/s horizontally, and a box of mass 4.5 kg is moving with a velocity of 4.0 m/s at a angle of 120 degrees from the horizontal. the two collide and stick together. Find the final velocity of the pair.

Homework Equations

Px = Pix = Pfx
MaVa + MbVb = Vf(Ma+Mb)

The Attempt at a Solution

My first question is whether or not, we take into account the angular velocity of the ball. With that said, I started off the solution using the above formula for x and y.

In the X:

MaVax + MbVbx = Vf(Ma+Mb)

with Vbx = 4 cos 120, this came out to .85ms-1, is this correct?

 

[cepheid]

Welcome to PF,

This is a 2D problem. You need to consider consv. of momentum in both the x and y directions. The problem asks for the final velocity of the objects, not just for the x-component of their final velocity.

EDIT: I haven't checked your arithmetic, but your method for the x-direction looks correct.

EDIT 2: What makes you think the ball has any angular velocity? In any case, this is not relevant to the problem.

 

[Squizzel]

Thanks for the welcome, and I understand what you are saying. I didn't do the Y part because I wanted to make sure I was doing the x correctly. So with that said, what do you think.

Also in the Y part, there is no velocity for the Ball, Ma.

 

[cepheid]

Thanks for the welcome, and I understand what you are saying. I didn't do the Y part because I wanted to make sure I was doing the x correctly. So with that said, what do you think.

Also in the Y part, there is no velocity for the Ball, Ma.

See the edits to my above post. Also, I do not get the same answer for the x-component of the velocity. Can you show your computational steps?

 

[Squizzel]

Sorry, I just noticed the v of the ball is 6, not 6.9 . But my calculation was this:

2.5 x 6 + 4.5 x 4 cos 120 = Vfx x 7
=.85

Is this the right angle? Because when I do the calculation for the Y, I get this:

Y = 4.5 x 4 cos 120 = Vfx x 7

= -1.28

This would mean that the velocity is negative.

 

[cepheid]

Sorry, I just noticed the v of the ball is 6, not 6.9 . But my calculation was this:

2.5 x 6 + 4.5 x 4 cos 120 = Vfx x 7
=.85

Is this the right angle? Because when I do the calculation for the Y, I get this:

Y = 4.5 x 4 cos 120 = Vfx x 7

= -1.28

This would mean that the velocity is negative.

Okay, the x-equation looks fine. For the y-equation: why are you using cosine for the y-component of the velocity?

You know that this answer cannot be right, because the vertical momentum is conserved, and the initial vertical momentum is all upward, which means that the final vertical momentum must also be upward (draw a picture).

 

[Squizzel]

Ok I now get Y = 4.5 x 4 sin 120 = Vfx x 7

Which comes out to 2.22 and a total v of 3.07. This seems correct. Thanks for your help.

 

[cepheid]

Ok I now get Y = 4.5 x 4 sin 120 = Vfx x 7

Which comes out to 2.22 and a total v of 3.07. This seems correct. Thanks for your help.

I agree with your y-component of the final velocity, but NOT with your total magnitude. Bear in mind that you compute the total velocity by taking a *vector sum* of the x and y-components, and that these are *perpendicular* to each other.

 

[Squizzel]

I see, I forgot about that.

I got 2.39 after taking the square root of both of them squared.

 

[cepheid]

I see, I forgot about that.

I got 2.39 after taking the square root of both of them squared.

Seems right to me. Always draw a diagram when doing these problems. It helps immensely, esp when it comes to avoiding errors like these.