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valentinak56 [21]
3 years ago
11

A 5.00-kg ball, moving to the right at a velocity of +2.00 m/s on a frictionless table, collides head-on with a stationary 7.50k

g ball. find the final velocities of the ball if the collision is elastic
Physics
1 answer:
valina [46]3 years ago
6 0
Using the law of conservation of momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂ v₂; u₂ = 0
5 x 2 = 5v₁ + 7v₂
10 = 5v₁ + 7v₂
v₂ = (10 - 5v₁)/7
Elastic collision so kinetic energy conserved:
1/2 x 5 x 2²  = 1/2 (5v₁² + 7v₂²)
20 = 5v₁² + 7((10 - 5v₁)/7)²
140 = 35v₁² + 5v₁² + 100 - 20v₁
40v₁² - 20v₁ - 40 = 0
v₁ = 1.28 m/s OR v₁ = -0.78 m/s
The first ball is lighter so its velocity will change direction due to collision.
v₁ = -0.78 m/s

10 = 5(-0.78) + 7v₂
v₂ = 1.98 m/s
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if the forces acting upon an object are balanced, then an object must a) be moving. b) be accelerating. c) be beginning to slow
Dennis_Churaev [7]

The correct answer to the question is : D) Be moving at a constant velocity.

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3 years ago
A machine can never be 100% efficient because some work is always lost due to what?
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When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

3 0
3 years ago
PLZ EXPLAIN IM SO CONFUSED AND THIS IS DUE TONIGHT. I WILL GIVE 50 POINTS!
bezimeni [28]

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I hope this helps. :)

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8 0
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