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sergij07 [2.7K]
4 years ago
8

Objects A and B each have a mass of 25 kilograms. Object A has a velocity of 5.98 meters/second. Object B is stationary. They un

dergo a perfectly elastic collision in one dimension. What is the total kinetic energy of the system after the collision?
A.)1.2 x 10^2 J
B.)4.5 × 10^2 joules
C. 5.0 × 10^2 joules
D. 9.5 × 10^2 joules
E. 1.1 × 10^3 joules
Physics
2 answers:
Vinil7 [7]4 years ago
6 0
There are numerous information's already given in the question. Based on those information's, the answer can be easily deduced.

Mass of the objects = 25 kg
Velocity of object A = 5.98 m/s
Velocity of object B = 0 m/s
Total kinetic energy = (1/2) * mass * (velocity)^2
                                = 0.5 * 25 * (5.98)^2
                                = 4.5 * 10^2 joules.
From the above deduction, we can conclude that the correct option among all the options that are given is option "B".
Genrish500 [490]4 years ago
4 0
<h3><u>Answer;</u></h3>

  = 4.5 * 10^2 joules.

<h3><u>Explanation;</u></h3>
  • A perfectly elastic collision is a type of collision in which kinetic energy is conserved.
  • Therefore, to calculate the kinetic energy of the system after the collision we use the formula 1/2 mv²

Mass = 25 kg

Velocity = 5.98 m/s

KE = 1/2 × 25 × 5.98 ²

     = 447 Joules

This translates to about <em><u>4.5 ×10^2 Joules</u></em>

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Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

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as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

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                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

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                                  φ = 0

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                             ⇒ v= A\omega_{o}

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<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

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\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

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Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

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