Question

A 1500-kg blue convertible is travelling south, and a 2000-kg red sports utility vehicle is travelling west. If the momentum of the system consisting of the two cars is 8000kg•m/s directed at 60° west of south, what is the speed of each car?

Answer 1

A 1500-kg blue convertible is travelling south, and a 2000-kg red sports utility vehicle is travelling west. If the momentum of the system consisting of the two cars is $8000\mathrm{kg}\cdot \mathrm{m} / \mathrm{s}$ directed at $60{}^{\circ}$ west of south, what is the speed of each car?

The momentum of the two car system can be found from equations:

$$\left\{ \begin{array}{c} (m _ {1} v _ {1}) ^ {2} + (m _ {2} v _ {2}) ^ {2} = p ^ {2} \\ \cos (6 0) = \frac {v _ {2}}{v} \end{array} \right.$$ $$\left\{ \begin{array}{c} (m _ {1} v _ {1}) ^ {2} + (m _ {2} v _ {2}) ^ {2} = p ^ {2} \\ \frac {1}{2} = \frac {(m _ {1} + m _ {2}) v _ {2}}{p} \end{array} \right.$$ $$v _ {2} = \frac {p}{2 (m _ {1} + m _ {2})} = \frac {8 0 0 0 k g m / s}{2 (1 5 0 0 k g + 2 0 0 0 k g)} \cong 1. 1 4 m / s$$

From the first equation:

$$v _ {1} = \frac {1}{m _ {1}} \sqrt {p ^ {2} - (m _ {2} v _ {2}) ^ {2}}$$ $$v _ {1} = \frac {1}{1 5 0 0 k g} \sqrt {(8 0 0 0 k g m / s) ^ {2} - (2 0 0 0 k g * 1 . 1 4 m / s) ^ {2}} \cong 5. 1 1 m / s$$

Answer: $v_{1} \cong 5.11m / s$ , $v_{2} \cong 1.14m / s$