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zmey [24]
2 years ago
9

In the oscillating spring ball system, where is the velocity of the ball the greatest?

Physics
1 answer:
nlexa [21]2 years ago
6 0

The speed of an object in a mass-spring system is given under the function

v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}

Here,

m = mass

k = Spring constant

A = Amplitude

x = Position

When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as

v_{max}= A\sqrt{\frac{x}{m}}

So the correct answer is B.

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
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Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

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v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

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