A 29000-kg open railroad car, initially coasting at 0.825 m/s with negligible friction, passes under a hopper that dumps 117500 kg of scrap metal into it.

a)What is the final speed, in meters per second, of the loaded freight car?
b)How much kinetic energy is lost, in joules, when the freight car receives this scrap metal? Ignore the free-fall kinetic energy of the scrap metal.

Answers

Answer 1

Answer:

a) Answer: 0.16331 m/s

b) Answer: 8522.66 joules

Explanation:

Answer 2

The final speed is  0.1633 m/s.

When the freight car receives this scrap metal, 7915.72 Joule kinetic energy is lost.

What is momentum?

A body has momentum while it is moving, according to our understanding. It is said that a body's momentum is equal to the sum of its mass and speed. A body's direction is important when discussing momentum. Its direction belongs to the body's direction of motion.

(a) Given parameter:

Mass of the railroad car, M = 2900 Kg.

Mass of the scrap metal, m= 11750 Kg.

Initial speed of the car, u = 0.0825 m/s.

Initial speed of the scrap metal = 0 m/s.

Final speed of the system, v = ?

Then, initial momentum of the  system = initial momentum of the car +  initial momentum of the scrap metal .

= (29000×0.825 + 117500×0) kg.m/s.

= 23925 kg.m/s.

Final momentum of the system = total mass of the system× final speed

= (29000 + 117500)× v kg.m/s

= 146500v kg.m/s.

Hence, from principle of conservation of momentum,

initial momentum of the  system = Final momentum of the system

⇒ 23925 = 146500v

⇒ v = 0.1633 m/s.

The final speed of the  loaded freight car is 0.1633 m/s.

(b) Lost in kinetic energy = initial kinetic energy of the car - final kinetic energy of the loaded car.

= 1/2*29000*0.825² -1/2*(29000 + 117500)* 0.1633² Joule.

= 7915.72 Joule.

Hence, 7915.72 Joule kinetic energy is lost in this process.

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I will give Brainliest answer and 5 stars
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___________________

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Answer:

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Please find attached photograph for your answer.

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just give the other person brainlyest

Explanation:

In a titration, 50.00 cm3 of 0.300 mol/dm3
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Answer:

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Dhdjrjfbjsjfnsmctejzbrnj

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Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi=0). (Figure 1)After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1>m2, so that after the collision, the two objects move off in the direction of the first object before the collision. What is the final speed vf of block 2?

Answers

The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:

           [tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]  

the momentum is defined by the product of the mass and the velocity of the body.

        p = mv

The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.

If the system is isolated, the forces during the collision are internal and the it  is conserved. Let's find the momentum is two instants.

Initial instant. Before crash.

      p₀ = m₁ u₀ + 0

Final moment. After crash.

      [tex]p_f = m_1 u_f + m_2 v_f[/tex]  

The momentum is preserved.

      p₀ = [tex]p_f[/tex]  

      [tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]  

Since the collision is elastic, the kinetic energy is conserved.

      K₀ = [tex]K_f[/tex]

      ½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex]  + ½ m₂  [tex]v_f^2[/tex]  

       

Let's write our system of equations.

       [tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]

       

Let's solve

       [tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]

       

       [tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]  

         

        [tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]

In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:

           [tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]  

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