# Three blocks colliding

1. Nov 3, 2009

### mmoadi

1. The problem statement, all variables and given/known data

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. Relevant 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?

2. Nov 3, 2009

### 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

3. Nov 4, 2009

### mmoadi

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?

4. Nov 4, 2009

### 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:

5. Nov 4, 2009

### mmoadi

This is how I approached the problem graphically.

File size:
107.6 KB
Views:
61
6. Nov 4, 2009

### 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.

7. Nov 4, 2009

### mmoadi

Here we go:

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

Last edited by a moderator: May 4, 2017
8. Nov 4, 2009

### Delphi51

Ah, that makes sense and your calcs are correct!

9. Nov 5, 2009

### mmoadi

Thank you for helping and have a nice day!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook