# one or two dimensions of collisions

• Nov 12th 2007, 12:43 PM
rcmango
one or two dimensions of collisions
Kevin has a mass of 78.5 kg and is skating with in-line skates.

He sees his 22.50 kg younger brother up ahead standing on the sidewalk,
with his back turned.

Coming up from behind, he grabs his brother and rolls off at a speed of 1.95 m/s.

Ignoring friction, find Kevin's speed just before he grabbed his brother.
in m/s

i'm not sure if this is for collisions in one or two dimensions, or both?
• Nov 12th 2007, 01:00 PM
Right cant remember my physics very well but pretty sure its just...

$\displaystyle m_1(v_1)^2 = m_2(v_2)^2$

which in his case is $\displaystyle 100(1.95^2)= 78.5{(v_2)^2}$

which gives $\displaystyle v_2 = 2.2m/s$
• Nov 12th 2007, 07:44 PM
rcmango
i got it.

i used (m1 + m2) / m1 * vf

and got 2.51 m/s

thanks.
• Nov 12th 2007, 07:45 PM
topsquark
Quote:

Originally Posted by rcmango
Kevin has a mass of 78.5 kg and is skating with in-line skates.

He sees his 22.50 kg younger brother up ahead standing on the sidewalk,
with his back turned.

Coming up from behind, he grabs his brother and rolls off at a speed of 1.95 m/s.

Ignoring friction, find Kevin's speed just before he grabbed his brother.
in m/s

i'm not sure if this is for collisions in one or two dimensions, or both?

Quote:

Right cant remember my physics very well but pretty sure its just...

$\displaystyle m_1(v_1)^2 = m_2(v_2)^2$

which in his case is $\displaystyle 100(1.95^2)= 78.5{(v_2)^2}$

which gives $\displaystyle v_2 = 2.2m/s$

We aren't told that energy is conserved, so we cannot assume it. Also, since this is an archetypal inelastic collision problem, I can guarantee that energy is not conserved.

So consider momentum conservation. There are two objects before the collision and (effectively) only one afterward. Set a coordinate system with +x in the direction that Kevin is initially moving in. At the collision point there are no net external forces acting on the two brothers, so momentum is conserved during the collision. So:
$\displaystyle P_{0, tot} = P_{tot}$

$\displaystyle m_1v_{01} + m2v_{02} = (m_1 + m_2)v_{12}$

Kevin's brother is not moving initially so we have that $\displaystyle v_{02} = 0~m/s$.

$\displaystyle m_1v_{01} = (m_1 + m_2)v_{12}$

$\displaystyle v_{01} = \frac{m_1 + m_2}{m_1} \cdot v_{12}$

Thus
$\displaystyle v_{01} = \frac{78.5 + 22.50}{78.5} \cdot 1.95 = 2.50892~m/s$

-Dan