Question

At the intersection of the Figure below, a yellow subcompact car with mass 950 kg
traveling east collides with a red pickup truck with mass 1900 kg that traveling north and
ran a red light. The two vehicles stick together as a result of the collision, and the
wreckage slides at 16.0 m/s in the direction 24.0o east of north y (north). Calculate the
speed of each vehicle before the collision. The collision occurs during a heavy rainstorm;
you can ignore friction.

Answer 1

Answer on Question #72188, Physics / Molecular Physics | Thermodynamics

Question. At the intersection of the Figure below, a yellow subcompact car with mass $950 \, kg$ traveling east collides with a red pickup truck with mass $1900 \, kg$ that traveling north and ran a red light. The two vehicles stick together as a result of the collision, and the wreckage slides at $16.0 \, m/s$ in the direction $24.0{}^\circ$ east of north y (north). Calculate the speed of each vehicle before the collision. The collision occurs during a heavy rainstorm; you can ignore friction.

Given. $\alpha = 24.0{}^{\circ}$ ; $v_{f} = 16.0 \, m/s$ ; $m_{1} = 950 \, kg$ ; $m_{2} = 1900 \, kg$ .

Find. $v_{1}, v_{2} - ?$ .

Solution.

According to the Law of Momentum Conservation, we have

$$m _ {1} \vec {v} _ {1} + m _ {1} \vec {v} _ {1} = (m _ {1} + m _ {2}) \vec {v} _ {f}$$

in projections on the axis

$$X: m _ {1} v _ {1} = (m _ {1} + m _ {2}) v _ {f x} = (m _ {1} + m _ {2}) v _ {f} \sin \alpha$$ $$Y: m _ {2} v _ {2} = (m _ {1} + m _ {2}) v _ {f y} = (m _ {1} + m _ {2}) v _ {f} \cos \alpha$$

Hence

$$v _ {1} = \frac {(m _ {1} + m _ {2}) v _ {f} \sin \alpha}{m _ {1}} = \frac {(9 5 0 + 1 9 0 0) \cdot 1 6 \cdot \sin 2 4 {}^ {\circ}}{9 5 0} = 1 9. 5 m / s$$ $$v _ {2} = \frac {(m _ {1} + m _ {2}) v _ {f} \cos \alpha}{m _ {2}} = \frac {(9 5 0 + 1 9 0 0) \cdot 1 6 \cdot \cos 2 4 {}^ {\circ}}{1 9 0 0} = 2 1. 9 m / s$$

Answer. $v_{1} = 19.5 \, m / s$ ; $v_{2} = 21.9 \, m / s$ .

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