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3 years ago

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Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen

t correctly describes the change in momentum of the two balls?
a. |ΔpBl<|ΔPA|
b. |ΔpBl=|ΔPA|
c. |ΔpB|>|ΔPA|
d. ΔpB > ΔPA

1 answer:

muminat3 years ago

4 0

Answer:

Option A

Explanation:

From the question we are told that:

Mass $m=0.20kg$

Velocity $v=4m/s$

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

$M_{a1}=mV$

$M_{a1}=0.2*4$

$M_{a1}=0.8$

Final Momentum

$M_{a2}=-0.8kgm/s$

Therefore

$\triangle M_a=-1.6kgm/s$

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

$M_{b1}=mV$

$M_{b1}=0.2*4$

$M_{b1}=0.8$

Final Momentum

$M_{b2}=-0 kgm/s$

Therefore

$|\triangle M_a|>|\triangle Mb|$

Option A

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Answer:

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2 years ago

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As we know that KE and PE is same at a given position

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also the PE is given as function of position as

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now it is given that

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now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

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Part b)

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we can use the formula

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now to find the fraction of kinetic energy

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$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

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now since total energy is sum of KE and PE

so fraction of PE at the same position will be

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3 years ago

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Explanation:

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solution

we know that

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here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

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