EK Physics - Section 3 - Q67 and Q68

[SKaminski]

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Section 3, The last set of lecture questions.

Q67:
Question:
Objects A and B are placed on the spring as shown. Object A has twice as much mass as object B. If the spring is depressed and released, propelling the objects into the air, object A will:

A) rise one fourth as high as object B.
B) rise half as high as object B.
C) rise to the same height as object B.
D) rise twice as high as object B.

My answer: B.
Correct Answer: C.

Their justification: An explination about taking the question to extremes (what if one weighted 1 million kg, and 1 weighted 1 kg, would the one weighing 1 kg go flying?) I think that's silly.

My justification: The equation for force from a spring is F=-k(change)x. Based on that, it's going to impart equal upward force onto each of them. Now, for the forces OPPOSING the upward force. F=ma. Acceleration is constant, so we only have to worry about.(downward)F=m. If Object's A has 2m, then its downward (opposing) force is twice that of object B, wouldn't that mean it rises twice half as high?

I know that gravity is suppose to work constant, but if THAT'S the explanation, does that mean that the force that gravity exerts is... vastly different given the objects?

Q68)
A spring powered dart-gun fires a dart 1 m vertically into the air. In order for the dart to go 4 m, the spring would have to be depressed:
A) 2 times the distance.
B) 3 times the distance.
C) 4 times the distance.
D) 8 times the distance.

Their answer: A
My answer: C

Their justification: 1/2Kx^2 = mgh
My justification: F = -k(change)x

It seems that we chose different problems. There's calculates height, mine calculates force, which is related to acceleration.

Maybe i'm not understanding acceleration well? What do you guys think.

 

[Yorick]

Q67: Hard to answer without seeing the picture.

Q68: This is a simple conservation of energy question. When the spring is compressed, let that distance be x and all of its energy is the potential energy of the spring or (1/2)kx^2. Now, when the dart is fired 1 m into the air, at that point it has a height h=1m and all of its energy is gravitational potential energy or mgh=mg(1)=mg (m is the mass of the dart. So our first equation is (1/2)kx^2=mg. Now, for the spring to fire the dart 4 m into the air, the height h=4 and the spring will be compressed a new distance and let that distance be y. So our equation now is (1/2)ky^2=mgh=4mg. From the first equation, mg=(1/2)kx^2 so substitute mg into the second equation to get (1/2)ky^2=4(1/2)kx^2, cross out like terms to get y^2=4x^2 and take the square root of both sides to get y=2x (the answer is positive because y and x are distances), so since x was the initial compressed distance and y=2x, the spring would now be compressed 2 times the distance.