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Chapter 6: Problem 50

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity \(13.0 \mathrm{~m} / \mathrm{s}\) toward the east, and the other is traveling north with speed \(v_{2 i}\). Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of \(55.0^{\circ}\) north of east. The speed limit for both roads is \(35 \mathrm{mi} / \mathrm{h}\), and the driver of the northward-moving vehicle claims he was within the limit when the collision occurred. Is he telling the truth?

Short Answer

Expert verified
No, the driver was not telling the truth as he was driving at approximately 41.54 mph, which is higher than the speed limit of 35 mph.

Step by step solution

01

Express Conservation of Momentum

Using the principle of conservation of momentum, the sum of the momentum of the two vehicles before the collision equals the momentum after the collision. Given \(m\) as the mass of both vehicles, velocity of first vehicle \(v_{1i} = 13 m/s\), unknown velocity of second vehicle \(v_{2i}\), and the velocity of vehicles after collision \(v_f\) moving at \(55^{\circ}\) (north of east), we express conservation of momentum in both \(x\) (east-west) and \(y\) (north-south) directions. In \(x\)-direction: \(m*v_{1i} = m*v_f*cos(55^{\circ})\). In \(y\)-direction: \(m*v_{2i} = m*v_f*sin(55^{\circ})\). As masses are same and non-zero, they cancel out in both equations.
02

Find Final Velocity

From equation from \(x\)-direction, we can find the final velocity \(v_f\). Solving \(v_{1i} = v_f*cos(55^{\circ})\) gives us \(v_f = v_{1i}/cos(55^{\circ})\). Substituting \(v_{1i} = 13 m/s\) and \(cos(55^{\circ}) = 0.5736\), we get \(v_f \approx 22.67 m/s\).
03

Find Initial Speed of Second Vehicle

Substitute the calculated final velocity \(v_f\) to the equation from \(y\)-direction to find the initial velocity of the second vehicle. Solving \(v_{2i} = v_f*sin(55^{\circ})\), and substituting \(v_f \approx 22.67 m/s\) and \(sin(55^{\circ}) = 0.8192\), we obtain \(v_{2i} \approx 18.58 m/s\).
04

Converting to mph and Comparing with Speed Limit

To check whether the driver of the northward-moving vehicle is telling the truth or not, we need to compare the calculated speed limit in mi/h. As \(1 m/s = 2.23694 mph\), the speed in mph of the second vehicle is \(v_{2i} * 2.23694 \approx 41.54 mph\). Compare this to the speed limit of 35 mph.

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Key Concepts

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

Collision Analysis
In this exercise, we delve into the fascinating realm of collision analysis. When two objects, like the automobiles in this scenario, collide, we can leverage the principles of physics to understand the outcomes of such events. Here, we also consider observations such as the angle at which the vehicles skid post-collision.

In our example, the vehicles collide and leave skid marks at an angle of 55 degrees north of east. This angle holds valuable clues about how the momentum of each vehicle combined to produce a motion in a new direction. It acts much like a vector that combines the individual momentums of the cars prior to collision.
  • The direction of the skid marks depends on the relative velocities of the vehicles before impact.
  • This analysis is vital as it helps recreate the scenario of the accident, providing insights into each vehicle鈥檚 motion leading up to the collision.
Velocity Calculation
Calculating velocities during the events of a collision is crucial for understanding motion dynamics. In this case, our goal is to find out how fast the northward-moving vehicle was traveling just before the crash.

Initially, we begin by figuring out the resultant velocity after the collision using the law of momentum conservation. The final velocity of the combined mass of both cars needs to be calculated first because it connects directly to the pre-collision speeds.
  • By observing the vehicle skidding at an angle of 55 degrees, we can separate this final motion into components along right-angled coordinates, usually depicted as the x (east-west) and y (north-south) axes.
  • Using trigonometric relationships, particularly the cosine and sine of the angle, we deduce the individual contributions of pre-collision speeds to the overall final velocity.
This systematic approach allows us to computationally reverse-engineer the conditions before impact, leading to the calculation of the initial velocity of the northward vehicle.
Momentum Conservation Equations
Momentum conservation serves as one of the most valuable tools for analyzing collisions. This law states that the total momentum of a closed system is conserved, which means it remains the same before and after a collision.

In our problem, the principle applies to both the x and y directions. By setting up equations for both directions, we can solve for unknowns such as the initial velocities of the vehicles:
  • In the x-direction, the momentum contributed by the eastward vehicle is equated to the post-collision velocity component along this line.
  • In the y-direction, the momentum associated with the northward vehicle matches its contribution to the final velocity's vertical component.
The beauty of these equations lies in their ability to fuse algebraic elements with physical observations, yielding meaningful results such as verifying if the northward driver鈥檚 claimed adherence to speed limits holds true. This elegant synthesis of math and physics illuminates the realities hidden in collision scenarios.

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Most popular questions from this chapter

Two objects of masses \(m\) and \(3 m\) are moving toward each other along the \(x\) -axis with the same initial speed \(v_{0}\). The object with mass \(m\) is traveling to the left, and the object with mass \(3 m\) is traveling to the right. They undergo an elastic glancing collision such that \(m\) is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two objects. (b) What is the angle \(\theta\) at which the object with mass \(3 m\) is scattered?

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block initially at rest on a horizontal surface. After impact, the block slides \(7.5 \mathrm{~m}\) before coming to rest. If the coefficient of kinetic friction between block and surface is \(0.650\), what was the speed of the bullet immediately before impact?

GP A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\) -direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second piece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\) -plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(m_{\mathrm{g}}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\) - and \(y\) -directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, v_{2}\), and \(v_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\) -components of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\) -components of the momenta of \(m_{1}\) and \(m_{2} .\) (e) Substitute the known momentum components into the general equations of momentum for the \(x\) - and \(y\) -directions, along with the known mass \(m_{3} .\) (f) Solve the two momentum equations for \(v_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(v_{3}\). (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain \(\tan \theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

A \(1200-\mathrm{kg}\) car traveling initially with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) in an easterly direction crashes into the rear end of a 9000 -kg truck moving in the same direction at \(20.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.42\) ). The velocity of the car right after the collision is \(18.0 \mathrm{~m} / \mathrm{s}\) to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss in energy.

ecp Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted onthe truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{~m} / \mathrm{s}\) and that they undergo a perfectly inelastic headon collision. Each driver has mass \(80.0 \mathrm{~kg} .\) Including the masses of the drivers, the total masses of the vehicles are \(800 \mathrm{~kg}\) for the car and \(4000 \mathrm{~kg}\) for the truck. If the collision time is \(0.120 \mathrm{~s}\), what force does the seat belt exert on each driver?
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