Physics Q.

[nothing123]

A mass m starting at point A is projected with the same initial
horizontal velocity v0 along each of the three tracks shown here
(with negligible friction) sufficient in each case to allow the mass
to reach the end of the track at point B. (Path 1 is directed up,
path 2 is directed horizontal, and path 3 is directed down.) The
masses remain in contact with the tracks throughout their
motions. The displacement A to B is the same in each case, and
the total path length of path 1 and 3 are equal. If t1, t2, and t3 are
the total travel times between A and B for paths 1, 2, and 3,
respectively, what is the relation among these times?

Picture attached.

These are the options;
a) t3<t2<t1
b) t2<t3<t1
c) t2<t1=t3
d) t2=t3<t1

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[bjb305]

A mass m starting at point A is projected with the same initial
horizontal velocity v0 along each of the three tracks shown here
(with negligible friction) sufficient in each case to allow the mass
to reach the end of the track at point B. (Path 1 is directed up,
path 2 is directed horizontal, and path 3 is directed down.) The
masses remain in contact with the tracks throughout their
motions. The displacement A to B is the same in each case, and
the total path length of path 1 and 3 are equal. If t1, t2, and t3 are
the total travel times between A and B for paths 1, 2, and 3,
respectively, what is the relation among these times?

Picture attached.

These are the options;
a) t3<t2<t1
b) t2<t3<t1
c) t2<t1=t3
d) t2=t3<t1

t3<t2<t1
reason being, for t3, you go downhill first, therefore picking up speed, then slowing back down to the original speed on the uphill
t2 stays at the original speed with no hills
t1 slows down from going uphill, then speeds up to come back to the original speed therefore taking the longest time

 

[BerkReviewTeach]

A mass m starting at point A is projected with the same initial
horizontal velocity v0 along each of the three tracks shown here
(with negligible friction) sufficient in each case to allow the mass
to reach the end of the track at point B. (Path 1 is directed up,
path 2 is directed horizontal, and path 3 is directed down.) The
masses remain in contact with the tracks throughout their
motions. The displacement A to B is the same in each case, and
the total path length of path 1 and 3 are equal. If t1, t2, and t3 are
the total travel times between A and B for paths 1, 2, and 3,
respectively, what is the relation among these times?

These are the options;
a) t3<t2<t1
b) t2<t3<t1
c) t2<t1=t3
d) t2=t3<t1

This is a really good conceptual question that unfortunately cannot be answered given their information.

The total time of travel is found by taking the distance traveled and dividing by the average speed. d = r x t, so t = d/r.

We know for sure that the average speed in path 1 is slowest and the average speed in path 3 is fastest.

We know this, because without friction, vA = vB, so a path that goes up at first (path 1) will go from vinitial to vslower back to vinitial at point B.

Likewise, in path 3, because it goes down at first, it goes from vinitial to vfaster back to vinitial at point B.

• This means that vavg Path 3 > vavg Path 2 > vavg Path 1.

The distance traveled in paths 1 and 3 is longer than in path 2.

As such, we know that in path 1 we have the longest distance and the slowest average speed, making t1 the largest amount of time. This only eliminates choice C.

Where the problem lies is with the vinitial. If vinitial is so amazingly large that the impact of acceleration on the average speed is negligible, then the shorter distance of path 2 than path 3 makes t2 shorter than t3. However, if the initial speed is so small that acceleration is substantial, then the average speed of path 3 exceeds that of path 2 by such a large amount that it more than compensates for the longer distance of path 3 than path 2. This would make t3 shorter than t2.

Get what you can from the concept, even though it's a poorly written question. Did their answer explanation address their error, because sometimes questions like this emphasize some test reality more than the science they are testing.

 

[Bleen]

I would go with C. They all begin with the same velocity and mass which means they all have the same initial kinetic energy. Since there is no friction they do not lose any energy. Both objects on path 1 and 3 experience a net force at some point (gravity). Object 1 loses kinetic energy equal to the potential energy it gains (remember, no friction so no net energy loss) when it ascends the hill, but gains that same kinetic energy back when it descends the hill. Object 3 does the same thing but in reverse; it gains kinetic energy then loses it. After the hill both objects have the same KE they started with. So, when they reach the the end of the path the KE final is equal to KE initial. Since KE is constant, and mass is constant, velocity must also be constant. If all objects have a constant velocity, then the time it takes to travel depends on the distance traveled. Object 2 has the shortest path, thus the shortest time, while objects 1 and 3 have the same longer path and thus a longer time. So t2 < t1 = t3.