A 2 500-kg car moving north collides with a 2 000-kg car moving west.After the collision,the cars stick together and move at an angle of 45° north of west at a speed of 8.0 m/s.What was the initial speed of the 2 500-kg car?

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Quizplus · Accepted Answer

To solve this problem, we use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. Since the cars stick together after the collision, we can treat them as a single object with a mass of $2,500 \, \text{kg} + 2,000 \, \text{kg} = 4,500 \, \text{kg}$.Let $v_{1}$ be the initial speed of the 2,500-kg car moving north and $v_{2}$ be the initial speed of the 2,000-kg car moving west. After the collision, the combined mass moves at a speed of $8.0 \, \text{m/s}$ at an angle of $45^\circ$ north of west.The momentum in the north direction before the collision (which comes only from the 2,500-kg car) is $2,500 \, \text{kg} \cdot v_{1}$. The momentum in the west direction before the collision (which comes only from the 2,000-kg car) is $2,000 \, \text{kg} \cdot v_{2}$.After the collision, the momentum can be broken down into north and west components based on the angle. Since the angle is $45^\circ$, the north and west components of the velocity are equal. The total momentum after the collision in the north and west directions is $4,500 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(45^\circ)$ for each direction.Setting the momentum in the north direction before and after the collision equal gives:$$2,500 \, \text{kg} \cdot v_{1} = 4,500 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(45^\circ)$$Solving for $v_{1}$, we use $\cos(45^\circ) = \frac{\sqrt{2}}{2}$:$$v_{1} = \frac{4,500 \cdot 8.0 \cdot \frac{\sqrt{2}}{2}}{2,500}$$$$v_{1} \approx 10.2 \, \text{m/s}$$Therefore, the initial speed of the 2,500-kg car was approximately 10.2 m/s.

A 2 500-Kg Car Moving North Collides with a 2