Question

001 10.0 points This question is typical on some driver's li- cense exams: A car moving at 45 km/h skids 13 m with locked brakes. How far will the car skid with locked brakes at 90 km/h? Assume that energy loss is due only to sliding friction Answer in units of m. 002 10.0 points Two identical objects are moving directly to- ward one another at the same speed v. 0 0 What is the total kinetic energy of the sys- tem of the two objects? 1. None of these 2. Still m u from the definition of kinetic 3. 4mu2 since (2v)2 4v2 2 energy 4. m ? since energy is a scalar 5. Zero since (7

Answer 1

Q1)

Work done by the brakes changes the car's kinetic energy from its initial value to zero when it finally comes to rest.

This work is equal to the force of friction causing the skid marks multiplied by the length of skid marks since -

Work = Force x Displacement

When the car's speed is increased from 45km/h to 90km/h, the velocity doubles. And as kinetic energy is -

KE = 0.5*m*v²

So, the kinetic energy gets increased by a factor of 4.

The force of friction between the road and the skidding tire doesn't change. So, the length over which the force is applied must increase 4 times as well. That means we would expect the skid marks to be almost 26m long.

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Q2)

Note that kinetic energy of individual objects = 0.5*m*v².

As you are including both the objects in the system, so the total energy of the system will be the sum of individual kinetic energy of each objects. And as energy is a scalar quantity, so the total energy of system would be = mv².

So, option (4) is correct.

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