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3 years ago

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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

as moving at 24.0 m/s to the east. After the explosion, the velocity of the m1 = 115 kg piece is 65.0 m/s to the east. Find the velocity (in m/s) (with a proper sign) of the other piece after the explosion

1 answer:

Daniel [21]3 years ago

5 0

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:

$Mv_i=m_1v_1+m_2v_2$

Which means (since we want $v_2$ and $M=m_1+m_2$ ):

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

So for our values we have:

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

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4 years ago

Two automobiles are 150 kilometers apart and traveling toward each other. One automobile

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3 years ago

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Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

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$\alpha = 0.714 \frac{rev}{s^{2} }$

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$\alpha = \frac{d \omega}{dt}$

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We know that from the equation of motion

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$\theta = \theta_1 + \theta_2$

$\theta =$ 17.5 + 35.01

$\theta =$ 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev

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