speed c/o physical science

Question

1.  a car is traveling north on a road at a speed of 10 meters per second.  it hits from behind another car which is at rest and the two cars stick together.  what can be concluded from this information?

 

a.  the collision was elastic.   I say no to answer a

b.  the two cars stuck together will move to the north at a speed less than 10 meters

c.  the two cars stuck together will move to the north at a speed more than 10 meters per second.  i think this is the answer

d.  momentum was not conserved in the collusion

 

2.  a car traveling 30 k/h suddenly speeds up to 60 k/h.  what can you conclude from this information?

 

a.  there was a change in the potential energy of the car  ( I don't think this is the answer since energy does not double)

b.  no work (in its scientific meaning) was done on the car

c.  the car will take four times longer to come to a complete stop

d.  all of these

 

 

Answer 1

Hey L,

 

Since the two cars are sticking together after the collision, this would be an inelastic case, which is given by the equation:

 

m1v1 + m2v2 = (m1 + m2)vf

 

In this case, lets treat m2 as the car at rest, this means that v2 = 0. If you plug in v1=10m/s, and we solve for the final velocity of the two cars stuck together, you will see that it has to be:

 

vf = (10 *m1)/(m1 + m2) 

 

which is less than 10m/s. The answer is b.

 

For the second question, if the velocity changes from 30 to 60 km/h, then we would have a change in kinetic energy, not potential energy. So answer a. is out of the question. For work to be done on the car, we need to apply a force on the car in the same direction as motion, but we aren't applying any forces on the car itself, so no work was done on the car. Answer b looks correct! But just to be sure, we can use kinematics to disprove answer c.:

 

vf1 = vi1 + at1    and    vf2 = vi2 + at2

 

Solving both equations for a, we can set the two equations equal to one another (since we are assuming acceleration will remain constant for both speeds). Keep in mind, the final velocity in either equation is zero, since we are coming to a complete stop!

 

a = -vi1/t1   and    a = -vi2/t2

 

 

-vi1/t1 = -vi2/t2

 

-30/t1 = -60/t2

 

30t2 = 60t1

 

t2 = 2t1

 

So the time it would take the car to stop when its going 60m/s is twice as long as when the speed is 30m/s. Answer c. cannot be right, and answer d. by default can't be correct if any of the answers are wrong.  Answer b. is the final!

Answer 2

1. The answer is (b).

 

Momentum is conserved, so the initial momentum equals the final momentum.  If the masses of the cars are m1 and m2, then:

 

m1*10 = (m1+m2)*v

 

v=10*m1/(m1+m2).  The fastest that the cars could be moving after the collision is 10 m/s, and that's if the mass of the car initially at rest has a mass of ZERO.  If the cars have equal masses, then the cars will move at 5 m/s after the collision.

 

2.  We can't conclude that there was a change in potential energy (I'm assuming they are talking about gravitational potential energy) because no information is given about the car's height.

 

The car will not take four times longer (in time) to stop.  The equation vf=vi+at shows that for a given acceleration, if velocity doubles, time doubles.

 

The Work-Kinetic Energy theorem states : Work = change in kinetic energy.  Since the kinetic energy has changed (KE depends on velocity: KE= 1/2mv^2), then there MUST have been work done on the car.

 

Now, if we look at distance, the car will take four times longer (in distance) to stop.  assuming a constant force, work is force x distance, and then work = change in KE.  Since the Kinetic Energy depends on the square of velocity, the kinetic energy of the car has quadrupled (increased by a factor of 4).  This means that the work required to stop the car is 4 times as much.  If we assume the same stopping force (friction from the tires), then this means the distance to stop must be 4 times as much.

 

The answer is (c)