What are the calculations for three blocks colliding and coming to rest?

[mmoadi]

Homework Statement

Two blocks (m1 = 0.02 kg, m2 = 0.03 kg, v1 = 1.5 m/s, v2 = 0.5 m/s) are sliding without friction on a surface. They are approaching each other at angle θ = 60º. In what direction and with how much velocity do we have to push the third block (m3 = 0.05 kg) against the first two blocks, so that when they crash they will come to rest?

2. Homework Equations

p=mv

3. The Attempt at a Solution

First two blocks:

G(x)= m(2)v(2)*(-sin θ)
G(y)= m(1)v(1) + m(2)*cos θ

Third block:

G(3x)= sin θ

G(3y)= m(1)v(1) – m(2)v(2)*cos θ

For the direction of the third block:

tan θ’= m(2)v(2)*sin θ / m(10v(1) + m(2)v(2)*cos θ → θ’= 19.1º

For the velocity of the third block:

G(3)= m(3)v(3) → v(3)= G3 / m(3)

G3= sqrt(m(1)²v(1)² + m(2)²v(2)² +2m(1)v(1)m(2)v(2)*cos θ)= 0.039686

v(3)= G3 / m(3)= 0.79372 m/s

Are my calculations correct?

 

[Delphi51]

I'm getting different answers. Did you get
G(x)= m(2)v(2)*(-sin θ) = .013
G(y)= m(1)v(1) - m(2)*cos θ = .0225

 

[mmoadi]

<sub>I retraced my steps and found out that I messed up big.

Here is how, I think, it is supposed to be:

First two blocks:

G(x)= m(1)v(1) + m(2)v(2) cos θ= 0.0375
G(y)= m(2)v(2) (-sin θ)= 0.013

Third block:

G(3(x))= -G(x)= -(m(1)v(1) + m(2)v(2) cos θ)= -0.0375
G(3(y))= -G(y)= m(2)v(2) sin θ= -0.013

For the direction of the third block:

tan θ’= m(2)v(2)*sin θ / m(10v(1) + m(2)v(2)*cos θ → θ’= 19.1º

For the velocity of the third block:

G(3)= m(3)v(3) → v(3)= G3 / m(3)

G3= sqrt(m(1)²v(1)² + m(2)²v(2)² +2m(1)v(1)m(2)v(2)*cos θ)= 0.039686

v(3)= G3 / m(3)= 0.79372 m/s

Are now my calculations correct?</sub>

 

[Delphi51]

There are several ways to look at this question! We'll have to agree on a diagram before we can understand each other. Show yours or use mine:

 

[mmoadi]

_{This is how I approached the problem graphically.}

 

[Delphi51]

It will be at least a few hours until your attachment is approved so I can see it.
An alternative is to upload the image to a free photo site such as photobucket.com and put a link to it here. If you can save the image as a jpg instead of bmp, it will be much smaller and faster to load.

 

[mmoadi]

Here we go:

http://www.slide.com/s/QrlHnyTY6D_vkTXmDIP5GFj8tT_LIBU0?referrer=hlnk

 

[Delphi51]

Ah, that makes sense and your calcs are correct!

 

[mmoadi]

Thank you for helping and have a nice day!