# More Collisions

When two object collide and stick together we say the collision is perfectly inelastic. In those situations the equations for conservation of momentum simplify a bit.

$$m_1u_1 + m_2u_2 = m_1v+m_2v$$ $$m_1u_1 + m_2u_2 = \left(m_1+m_2\right)v$$

Objects can also start together and separate, like in an explosion.

$$\left(m_1+m_2\right)u = m_1v_1 + m_2v_2$$
Example: A 100.0kg goalie throws a 1.1kg soccer ball while jumping forward at 3.2m/s. If the ball flies out of the goalie's hands at 10.0m/s what speed is the goalie?
solution

initial momentum = final momentum

(goalie and ball) = goalie + ball

$$(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}$$ $$(100.0+1.1)(3.2)=(100.0)v_{1}+(1.1)(10.0)$$ $$323.52=(100.0)v_{1}+11$$ $$3.13\small\frac{m}{s}=\normalsize v_{1}$$

# 2-D Conservation of Momentum

Momentum is a vector. It has a direction and a magnitude. We can solve for the horizontal and vertical components separately just like how we solved 2-d Motion problems. Although in this case there isn't a time variable to link up the vertical and horizontal equations.

Horizontal: A pink hexagon and a cyan triangle collide. Before the collision the hexagon has a mass of 5.20kg and is moving to the right at 2m/s. The cyan triangle is moving to the left at 1m/s and has a mass of 1.17kg. If the triangle bounces off the hexagon at 2.92 m/s to the right what is the horizontal velocity of the hexagon? (ignore the vertical information)
solution

initial momentum = final momentum

hexagon + triangle = hexagon + triangle

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(2)+(1.17)(-1)=(5.20)v_{1}+(1.17)(2.92)$$ $$10.40-1.17=(5.20)v_{1}+3.42$$ $$5.81=(5.20)v_{1}$$ $$1.12 \small\frac{m}{s}=\normalsize v_{1}$$

Vertical: Lets take a look at the vertical aspect of the pink hexagon and cyan triangle collision. Before the collision the hexagon has a mass of 5.20kg and is not moving up or down. The final vertical velocity of the hexagon is 0.23 down. The final vertical velocity of the triangle is 0.67 up. What is the triangle's initial vertical velocity?
solution

initial momentum = final momentum

hexagon + triangle = hexagon + triangle

$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(0)+(1.17)(u_{2})=(5.20)(-0.23)+(1.17)(0.67)$$ $$(1.17)(u_{2})=-0.41$$ $$u_{2}=-0.35\small\frac{m}{s}$$