AAMC Physics Self Assessment #22

[moto_za]

Need help with this question please. I thought decreasing g would decrease the time? How do you derive the appropriate formula for this question?

Q: Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in time t. If the value of g were reduced to g/6 (as on the moon) then t would:

A: increase by a factor of 6

TIA!

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[Captain Sisko]

Need help with this question please. I thought decreasing g would decrease the time? How do you derive the appropriate formula for this question?

Q: Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in time t. If the value of g were reduced to g/6 (as on the moon) then t would:

A: increase by a factor of 6

TIA!

v = v0 + a*t

where v is zero, since at the apex vertical velocity has zero magnitude. a is negative because gravity is pulling you down. t is really half the total time of flight, again because you're going to the apex.

so 0 = v - g*t/2

t = 2*v/g

so if g is now g/6

t = 2*v/(g/6) = 12*v/g

or six times the previous expression.

hope that helps!

 

[moto_za]

^ Thanks a lot!! Thinking about the question qualitatively, wouldn't it make sense for the time to decrease since there is less gravity?

 

[sazerac]

^ Thanks a lot!! Thinking about the question qualitatively, wouldn't it make sense for the time to decrease since there is less gravity?

Qualitatively, if g was enormously huge, the ball would return near instantaneous.

If g were ridiculously small, you could throw the ball up, and then have time to make some coffee and play a couple video games before the thing fell back to earth.

So, no, if there is less gravity I would think the time would increase.

 

[Postictal Raiden]

Need help with this question please. I thought decreasing g would decrease the time? How do you derive the appropriate formula for this question?

Q: Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in time t. If the value of g were reduced to g/6 (as on the moon) then t would:

A: increase by a factor of 6

TIA!

A very easy way to see the relationship is by taking a look at acceleration.

Acceleration = velocity/time

Now, you can see that the relationship between acceleration and time is inversely and linearly proportional. Meaning, if you decrease on my factor of x, the other will increase by factor of x.

I'm interested in knowing the answer choices given, could you provide us with them? It's because for a split second I thought that the answer should be square root of 6 because of this equation: x = 0.5*a*t^2. Therefore, I think if I were under time constraints and had to make a quick call, I would have gone with sqrt(6).

 

[milski]

A very easy way to see the relationship is by taking a look at acceleration.

Acceleration = velocity/time

Now, you can see that the relationship between acceleration and time is inversely and linearly proportional. Meaning, if you decrease on my factor of x, the other will increase by factor of x.

I'm interested in knowing the answer choices given, could you provide us with them? It's because for a split second I thought that the answer should be square root of 6 because of this equation: x = 0.5*a*t^2. Therefore, I think if I were under time constraints and had to make a quick call, I would have gone with sqrt(6).

That would be the case if the object was being dropped from a given height. In the problem here, the top point is determined by v=0 and since v is proportional to time, the factor stays the same.

 

[GomerPyle]

That would be the case if the object was being dropped from a given height. In the problem here, the top point is determined by v=0 and since v is proportional to time, the factor stays the same.

Wait, so when do you use x=.5gt^2? At the top point, essentially it's the same as dropping the object from that same height....

 

[Temperature101]

Wait, so when do you use x=.5gt^2? At the top point, essentially it's the same as dropping the object from that same height....

When the initial velovity is ZERO... This problem has an initial velocity that is not ZERO.

 

[GomerPyle]

When the initial velovity is ZERO... This problem has an initial velocity that is not ZERO.

Can't you start at the top of the parabolic motion where the ball has a velocity of zero, and start from there? Then you could use x=.5gt^2 but the answer would be different..

 

[sazerac]

Can't you start at the top of the parabolic motion where the ball has a velocity of zero, and start from there? Then you could use x=.5gt^2 but the answer would be different..

More importantly the X equation solves for the location of the ball (for instance the height at the top of the motion). But the question never asked you to compare the two max heights. It asked you to compare the two time intervals.


This is a great question because it might trick the reader into using X=0.5at^2 when V0 is not zero. The correct equation, as you will recall, is really
X=X0 + V0t + 0.5at^2

Since you don't know V0 and you don't know T, this really isn't the correct equation to use.