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Blizzard [7]
3 years ago
15

A father (75 kg) was standing watching TV, minding his own business when one of his kids (20 kg) approached him at 2m/s heading

at 0 degrees and jumped onto his back. Another kid (15 kg) approached him at 3m/s heading at 45 degrees to the first kid and also jumped on him at exactly the same time. Immediately after impact, in what direction did the group go (measured in degrees with respect to the first kid)
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

Explanation:

Conservation of momentum

Let's say the first child was headed north

North momentum

75(0) + 20(2) + 15(3sin45) = (75 + 20 + 15)vn

vn = 0.6529 m/s

East momentum

75(0) + 20(0) + 15(3cos45) = (75 + 20 + 15)ve

ve = 0.28927 m/s

θ = arctan(0.28927/0.6529)

θ = 23.896...

θ = 24° east of north

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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
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Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

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Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

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speed of the toy, v = 0.4 m/s

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E = 0.0416 J

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A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

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E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

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Therefore,

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<span>Now applying Newton's 2nd law of motion,
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<span>F = ma
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