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kompoz [17]
3 years ago
15

A 3.5 kg block slides along a frictionless horizontal with a speed of 2.5 m/s. After sliding a distance of 5 m, the block makes

a smooth transition to a frictionless ramp inclined at an angle of 50â—¦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 .
Physics
1 answer:
Andru [333]3 years ago
3 0

Answer:

  L = 0.416 m

Explanation:

In this exercise to calculate the distance traveled in the branch that climbs the block we can use the concepts of energy

Starting point Flat part

       Em₀ = K = ½ m v²

Final point. Highest part of the ramp

       Em_{f} = U = m g h

As there is no friction the mechanical energy is conserved

        Em₀ =  Em_{f}

        ½ m v² = m g h

       h = v² / 2g

       h = 2.5 2/2 9.8

       h = 0.3189 m

The distance traveled can be found with trigonometry

      sin 50 = h / L

      L = h / sin 50

      L = 0.3189 / sin50

      L = 0.416 m

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Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s

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\dot{V}=0.0733 \,m^3.s^{-1}

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