TBR Physics Section 1 Passage Questions

[capnamerica]

I'm having some difficulty understanding the answers to some of the passage questions in TBR Physics Section 1.

Q9.) Why is the answer D? I still don't understand this...If acceleration remains constant, shouldn't speed keep increasing as you go down the ramp?

Q27.) Looking at the solution TBR provides, how does one visualize the path off the projectile for the various angles? Of course, I can visualize that at an angle less than 45, the projectile has a longer range, whereas at an angle greater than 45, the projectile has a larger maximum height. However, I don't know how someone can visualize the more specific differences in trajectory due to small differences in the angle...Any advice!?

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[kentavr]

I'm having some difficulty understanding the answers to some of the passage questions in TBR Physics Section 1.

Q9.) Why is the answer D? I still don't understand this...If acceleration remains constant, shouldn't speed keep increasing as you go down the ramp?

Q27.) Looking at the solution TBR provides, how does one visualize the path off the projectile for the various angles? Of course, I can visualize that at an angle less than 45, the projectile has a longer range, whereas at an angle greater than 45, the projectile has a larger maximum height. However, I don't know how someone can visualize the more specific differences in trajectory due to small differences in the angle...Any advice!?

Q9. Learn this formula, it is very useful V^2 - (V_0)^2 = 2as. You can derive it if you want. I'll skip it. Since your initial speed V_0=0, then V^2=2as or
V=sqrt(2as) .
So the chart should be square root of s. It is D.

Q27. If you can distinguish between range for <45 and >45 that's all they need. To get a more detail view will be more complicated. There is a general equation that describes the parabola for free fall, but it is rather complicated to expect it on MCAT.
Just for fun
y=x*tg(alpha) - (1+tg^2(alpha))*g*x^2/(2*v_0)
So, as you can see, visualization for general case looks impossible (at least for me )

 

[capnamerica]

Q9. Learn this formula, it is very useful V^2 - (V_0)^2 = 2as. You can derive it if you want. I'll skip it. Since your initial speed V_0=0, then V^2=2as or
V=sqrt(2as) .
So the chart should be square root of s. It is D.

Q27. If you can distinguish between range for <45 and >45 that's all they need. To get a more detail view will be more complicated. There is a general equation that describes the parabola for free fall, but it is rather complicated to expect it on MCAT.
Just for fun
y=x*tg(alpha) - (1+tg^2(alpha))*g*x^2/(2*v_0)
So, as you can see, visualization for general case looks impossible (at least for me )

So then, how would you have solved this question without visualizing the details?

 

[kentavr]

So then, how would you have solved this question without visualizing the details?

I am sorry, I thought you got it and just need more details about the whole thing. Here is how I would think about it:

The question is: where the maximum range can be achieved? Below 45 degree or above? (Since that is the difference in pictures)

To every trajectory below 45 degrees there is a trajectory above 45 degrees that intersects it at the X- level of the hill. (Agree?)

Then at this intersection point the higher trajectory ball has the same absolute speed as lower trajectory ball(conservation of energy).

However, high ball has larger vertical speed (pointing down) and smaller horizontal speed. So it hit the ground faster and make less horizontal range.

That means that for every high trajectory there is a low trajectory with larger range and the answer to the question become that the maximum range can be achieved somewhere where trajectory angle is less then 45 degree.

There is only one chart that has this property.

Let me know, if it still unclear I will attach the picture.

P.S. Here it is