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 100.0 kg goalie throws a 1.1 kg soccer ball while jumping forward at 3.2 m/s. If the ball flies out of the goalie's hands at 10.0 m/s what speed is the goalie?
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{goalie holding ball = goalie + ball}$$ $$(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}$$ $$(100.0+1.1)(3.2)=(100.0)v_{1}+(1.1)(10.0)$$ $$323.52=(100.0)v_{1}+11$$ $$3.13 \, \mathrm{\tfrac{m}{s}} = v_{1}$$

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 5.20 kg and is moving to the right at 2 m/s. The cyan triangle is moving to the left at 1 m/s and has a mass of 1.17 kg. 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 $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{hexagon + triangle = hexagon + triangle}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(2)+(1.17)(-1)=(5.20)v_{1}+(1.17)(2.92)$$ $$10.40-1.17=(5.20)v_{1}+3.42$$ $$5.81=(5.20)v_{1}$$ $$1.12 \mathrm{\tfrac{m}{s}}= v_{1}$$

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 5.20 kg and is not moving up or down. The final vertical velocity of the hexagon is 0.23 m/s down. The final vertical velocity of the triangle is 0.67 m/s up. What is the triangle's initial vertical velocity?
solution $$\sum p_{\mathrm{initial}} = \sum p_{\mathrm{final}}$$ $$\text{hexagon + triangle = hexagon + triangle}$$ $$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$ $$(5.20)(0)+(1.17)(u_{2})=(5.20)(-0.23)+(1.17)(0.67)$$ $$(1.17)(u_{2})=-0.41$$ $$u_{2}=-0.35 \mathrm{\tfrac{m}{s}}$$
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