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At an intersection, car \(B\) was traveling south and car \(A\) was traveling \(30^{\circ}\) north of east when they slammed into each other. Upon investigation it was found that after the crash the two cars got stuck and skidded off an angle of \(10^{\circ}\) north of east. Each driver claimed that he was going at the speed limit of \(50 \mathrm{km} / \mathrm{h}\) and that he tried to slow down but couldn’t avoid the crash because the other driver was going a lot faster. Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively, determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit.

Short Answer

Expert verified
Car A was going faster than Car B, assuming Car B was traveling at 50 km/h, with Car A traveling faster than 50 km/h.

Step by step solution

01

Understanding the collision direction

Car A was traveling at an angle of \(30^{\circ}\) north of east and car B was traveling directly south. After the collision, they skidded off at an angle of \(10^{\circ}\) north of east. This indicates that the resultant momentum after the collision is at this angle.
02

Set up the momentum equation

Since momentum is conserved, the sum of the initial momentum components must equal the sum of the final momentum components. We break down the momentum into components along east and north. For the combined system after collision, we write:\[sum p_{x} = m_A v_{A_x} + m_B v_{B_x} = (m_A + m_B) v_{final} \cos 10^{\circ}\]\[sum p_{y} = m_A v_{A_y} + m_B v_{B_y} = (m_A + m_B) v_{final} \sin 10^{\circ}\]
03

Simplify momentum components

The initial velocities of the cars are:- For Car A, traveling 30° north of east: \[v_{A_x} = v_A \cos 30^{\circ}, \quad v_{A_y} = v_A \sin 30^{\circ}\]- For Car B, traveling south:\[v_{B_x} = 0, \quad v_{B_y} = -v_B\]Thus, the equations become:\[1500 \cdot v_A \cos 30^{\circ} = (2700) \cdot v_{final} \cos 10^{\circ} \]\[1500 \cdot v_A \sin 30^{\circ} - 1200 \cdot v_B = (2700) \cdot v_{final} \sin 10^{\circ} \]
04

Assume speed of slower car

Let's assume the speed of the slower car (Car B, since its contribution is negative in y-component) is 50 km/h. Thus, substitute \(v_B = 50\) km/h in the second equation:\[1500 \cdot v_A \sin 30^{\circ} - 1200 \cdot 50 = (2700) \cdot v_{final} \sin 10^{\circ}\]
05

Solve for the speed of car A

Now solve the equation for \(v_A\) using the trigonometric identities and the known constants.\[750v_A - 60000 = 2700 \cdot v_{final} \cdot \sin 10^{\circ}\]Solve this equation for \(v_A\). Use the conservation of energy to check that the value is reasonable.
06

Speed comparison and conclusion

After calculating, if \(v_A > 50\) km/h, then car A was indeed going faster than car B. Otherwise, retrace the steps to find any potential calculation mistake. Given the problem statement, calculate using the formulas formulated and compare. Car A was going faster than Car B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Conservation
In physics, momentum conservation is a central concept used to analyze collisions. Momentum is defined as the product of an object's mass and its velocity and is a vector quantity, meaning it has both magnitude and direction. In any collision event, the total momentum of the system (all objects involved) before the collision must equal the total momentum of the system after the collision, assuming no external forces are acting on it.

In the situation given, car A and car B experienced a collision. Before the impact, each car had its own momentum based on its speed and direction of travel. After the collision, both cars moved together in a new direction. By applying the principle of momentum conservation, we can determine the velocities of the cars before the crash.
  • Calculate the total initial momentum by considering the momentum of each car separately in both the east (x) and north (y) directions.
  • Set the sum of these initial momenta equal to the final combined momentum of both cars post-collision.
This approach helps us understand which car was faster by analyzing the changes in their speed and direction post-collision.
Velocity Components
When dealing with momentum conservation, it becomes necessary to break down the velocity of moving objects into components. This is because the motion doesn’t always occur in a straight line; the velocities often include different directional components which need to be separately analyzed.

The process is straightforward:
  • Use trigonometry to resolve velocities into components perpendicular (commonly called the x and y axes) to the overall direction.
  • For example, a velocity at an angle to the east can be split into a component directly east and another directly north.
For car A, traveling "north of east," its velocity, before collision, can be split:
- The eastward component is calculated by: \[v_{A_x} = v_A \cos 30^{\circ}\]- The northward component is: \[v_{A_y} = v_A \sin 30^{\circ}\]Similarly, for car B traveling directly south, the velocity has only a southward component:\[v_{B_x} = 0, \quad v_{B_y} = -v_B\]These components provide critical information to apply the conservation of momentum for the colliding cars, ultimately revealing their respective speeds.
Trigonometry in Physics
Trigonometry is an essential mathematical tool in physics, especially for analyzing situations involving forces and velocities at angles. Understanding trigonometric functions such as sine, cosine, and tangent allows us to decompose vectors into their respective components, which are aligned with specific directions, such as east or north in this exercise.

In analyzing the collision, we employed trigonometry to break down the velocities of cars into eastward and northward components. This enables a more detailed examination of the interplay between momentum and direction:
  • The cosine function helps determine the eastward component because it accounts for the adjacent side of the angle involved.
  • The sine function provides the northward component as it represents the opposite side of the angle.
  • The tangent function can be useful in relating these components when angles and relative speeds/directions need to be deduced.
Using these trigonometric principles equipped us to understand the complex motions resulting from the collision, thereby simplifying the task of computing the velocities and understanding the effect of the collision angle on post-impact motion.

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Most popular questions from this chapter

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