examkrackers physics 1001 questions 327 and 328 and 330

[2010premed]

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This is really confusing me... a bit of clarification is greatly appreciated

327: A 2kg ball and an 8kg ball are placed on separate springs, each with the same spring constant. The springs are compressed by the same distance and released. Which of the following is true about the velocities of the balls upon leacing the springs?
A. The 2kg ball has a velocity 4 times greater
B. The 2 kg ball has a v 2 times greater
C. Their v's are the same
D. The 8kg ball has a v 4 times greater
Answer: B


328. A 2kg ball and an 8kg ball are placed on the same spring at the same time. The spring is compressed and released. Which of the following is true about the maximum heights reached by the balls?
A. The 2kg ball will go four times as high.
B. The 2 kg ball will go twice as high
C. The balls will reach equal maximum heights
D. The 8kg ball will go 4 times as high
Answer: C

Ok this one is separate, but is also confusing me

330. A man puts springs on the bottoms of his shoes and jumps off a 10 m cliff. Assuming the springs follow Hooke's law, from the moment the springs touch the ground to when they are fully compressed, the magnitude of the man's acceleration will:
A. decrease the increase
B. increase then decrease
C. remain constant at 10m/s^2
D. remain constant at zero
Answer: A

 

[Phantastic]

This is really confusing me... a bit of clarification is greatly appreciated

327: A 2kg ball and an 8kg ball are placed on separate springs, each with the same spring constant. The springs are compressed by the same distance and released. Which of the following is true about the velocities of the balls upon leacing the springs?
A. The 2kg ball has a velocity 4 times greater
B. The 2 kg ball has a v 2 times greater
C. Their v's are the same
D. The 8kg ball has a v 4 times greater
Answer: B

If they are compressed to the same distance, according to F=-kx=ma, the one with less mass will have a greater magnitude of acceleration. That rules out C/D. Since Hooke's law is conservative: PE(i) + KE(i) = PE(f) + KE(f), and they both start with all PE, and end with all KE when leaving the spring--also, they both start with the same PE.

PE(i) = KE(f) = 0.5mv^2 ...as you can see, mass is inversely proportional to the square of the velocity. So if you have one fourth the mass, you won't have 4x the velocity...you will only have 2x the velocity.

These are two important concepts for conservative forces: The mass of an object is inversely proportional to the acceleration and the square of the velocity.

328. A 2kg ball and an 8kg ball are placed on the same spring at the same time. The spring is compressed and released. Which of the following is true about the maximum heights reached by the balls?
A. The 2kg ball will go four times as high.
B. The 2 kg ball will go twice as high
C. The balls will reach equal maximum heights
D. The 8kg ball will go 4 times as high
Answer: C

Notice that this time, they weren't compressed to the same distance. In fact, the only thing compressing the spring F = -kx = mg (the weight). Notice that x (the distance displaced) and m (the mass) are directly proportional. Again, we go from all PE to all KE, so: PE(i) = KE(f),

hmm..I forget the math here, but qualitatively the one with the greater force will also have the greater acceleration downward upon "liftoff" and will only go as high as the lighter one...pretty sure it's energy conservation.

Ok this one is separate, but is also confusing me

330. A man puts springs on the bottoms of his shoes and jumps off a 10 m cliff. Assuming the springs follow Hooke's law, from the moment the springs touch the ground to when they are fully compressed, the magnitude of the man's acceleration will:
A. decrease then increase
B. increase then decrease
C. remain constant at 10m/s^2
D. remain constant at zero
Answer: A

As the spring touch the ground, sum your forces: F(net_y) = F(spring) - mg

(the restoring force will "restore" in the positive y direction as it's being compressed, so I considered the force acting in the positive y direction).

So, the acceleration is decreasing at first, because F(spring) grows according to Hooke's law (directly proportional to the distance compressed), and mg stays the same at the surface of the earth. That's enough to go with A.

 

[Phantastic]

On the last example, it's worth noting that -mg is larger than F(spring), so as F(spring) grows all the way to the amplitude, it is only contributing to the decrease in the acceleration in the negative y direction. I probably should have made the downward y direction positive since that was the vector of my initial acceleration...

 

[2010premed]

Thanks a lot. Would you be to define when 2 diff masses would reach the same height? Does this happen with all spring problems?

 

[soundsofscience]

Hi,

I guess this is from a long time ago, but I got stuck on #328 in EK 1001 Physics as well.

I understand that both objects are now on the same spring. Because of this greater total weight of the two balls (rather than just one) pressing down on the spring, it compresses further. This greater Δx results in the storage of more elastic potential energy, which will eventually be converted into the kinetic energy that launches the balls.

The answer says that upon release, each ball leaves the spring with the same velocity ("because the balls leave the spring at the same time"). I am not sure that I understand why leaving the spring at the same time implies having the same velocity at release from the spring.

What would I need to know that would allow me to notice this?

The first thing that came to mind for me was to write a formula attached below. But I didn't know what to do with it, or if it was relevant, mainly because I didn't know how to divvy up the kinetic energy between the two objects.

The only thing I can think of is that if you knew that both balls reached the same peak height, and you knew that they left the spring at the same time, then by v = d/t, where d is height, you would know that they started with the same velocity. But you don't know beforehand that they reach the same height, as that is what the question is about.

Thanks!

 

[DrknoSDN]

Hi,

I guess this is from a long time ago, but I got stuck on #328 in EK 1001 Physics as well.

I understand that both objects are now on the same spring. Because of this greater total weight of the two balls (rather than just one) pressing down on the spring, it compresses further. This greater Δx results in the storage of more elastic potential energy, which will eventually be converted into the kinetic energy that launches the balls.

The answer says that upon release, each ball leaves the spring with the same velocity ("because the balls leave the spring at the same time"). I am not sure that I understand why leaving the spring at the same time implies having the same velocity at release from the spring.

What would I need to know that would allow me to notice this?

The first thing that came to mind for me was to write a formula attached below. But I didn't know what to do with it, or if it was relevant, mainly because I didn't know how to divvy up the kinetic energy between the two objects.

The only thing I can think of is that if you knew that both balls reached the same peak height, and you knew that they left the spring at the same time, then by v = d/t, where d is height, you would know that they started with the same velocity. But you don't know beforehand that they reach the same height, as that is what the question is about.

Thanks!

I would suggest thinking about this conceptually instead of using a formula. The spring is essentially a platform that is moving up (like an elevator), once the spring reaches neutral position the balls will both be moving the same speed because they were both on the same platform. If two objects are launched into the air the maximum height is not dependent on the mass of the object. Velocity is the only factor.

I visualized this question wrong when initially reading it. Thinking about 2 different mass balls in a catapult is probably easier to see they reach the same height.