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geniusboy [140]
3 years ago
12

In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f

ragment shot directly south at 9 m/s. What would be the x-component of the velocity of the m2 fragment after the explosion
Physics
1 answer:
Inga [223]3 years ago
6 0

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2  \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

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You can reason it out like this:

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-- Distance covered in 30 sec at an average speed of 120 m/s

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The formula that has all of this in it is the formula for
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                       =      (4 m/s²)          ·      (900 sec²)

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When you translate these numbers into units for which
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When the light turns green, Andy mashes the pedal to the metal
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He does zero to 60 mph in 3.4 seconds, and at the end
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He doesn't need to worry about getting a speeding ticket.
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3 0
3 years ago
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\huge \bold \blue{ \underline{ answer}}

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