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cluponka [151]
3 years ago
7

A 3.00 kg pool ball is moving to the left with a speed of 4.30 m/s without friction. If it experiences an impulse of -4.00 Ns, w

hat is the object's speed and direction after the impulse occurs?
Physics
1 answer:
brilliants [131]3 years ago
3 0

Explanation:

Given that,

Mass of thee pool ball, m = 3 kg

Initial speed of the ball, u = -4.3 m/s

Impulse experienced by thee ball, J = -4 N-s

To find,

Speed of the object after impulse occurs and its direction.

Solution,

Let left side is negative and right side is positive. So, the change in momentum or the impulse is given by the following expression as :

J=m(v-u)

v=\dfrac{J}{m}+u

v=\dfrac{-4}{3}+(-4.33)

v = -5.663 m/s

So, the speed of the object is 5.663 m/s and it is towards left.

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x(t=3s) = 0.07 m to the nearest hundredth

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Integration by parts is done this way...

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Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

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Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

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u = 2t

(du/dt) = 2

du = 2 dt

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v = (e⁻³ᵗ/9)

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Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

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