Block 1, of mass m1=2.50 kg, moves along a frictionless air track with speed v1=27.0m/s. It collides with block 2, of mass m2=21.0 kg, which was initially at rest. The blocks stick together after the collision.

A: Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.

B: Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.

C: what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

The momentum conservation  allows to find the results for the questions about the collision of the two particles are:

    A) The total moment is:  = 67.5 N m / s

    B) The system velocity after the collision is: v = 2.87 m / s

    C) The decrease in kinetic energy is: ΔK = - 814.47 J

Given parameters

To find

    A) The total moment.

    B) The final speed.

    C) The change in kinetic energy.

The momentum is defined by the product of the mass and the velocity, if a system is isolated in total momentum it is conserved, this is one of the most important quantities in physics.

         p = mv

Where the bold letters indicate vectors, m is the mass and v the velocity of the body.

We define the system formed by the two particles, therefore the forces during the collision are internal and the momentum is conserved.

Initial instant. Before the crash.

    p₀ = m₁ v₁ + 0

Final instant. After the crash.

    = (m₁ + m₂) v

a) The total momentum is

    = 2.5 27

     = 67.5 N m / s

b) the momentum is preserved.

    m₁ v₁ = (m₁ + m₂) v

    v =

Let's calculate.

     v =  

     v = 2.87 m / s

C) kinetic energy is the energy of motion.

          K = ½ m v²

we look for the initial and final kinetic energy.

          K₀ = ½ m1 v1²

           = ½ (m1 + m2) v²

The loss of kinetic energy is

          ΔK = K_f - K₀

          ΔK = ½ (m₁ + m₂) v² - ½ m₁ v₁²

Let's calculate

          ΔK = ½ (2.5 +21) 2.87² - ½ 2.5 27²

          ΔK = 96.78 - 911.25

          ΔK = - 814.47 J

The negative sign indicates that the energy decreases, this quantity is converted into internal energy of the system during the crash.

In conclusion using the conservation of momentum we can find the results for the questions about the collision of the two particles are:

    A) The total moment is:  = 67.5 N m / s

    B) The system velocity after the collision is: v = 2.87 m / s

    C) The decrease in kinetic energy is: ΔK = - 814.47 J

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A) total initial momentum = 67kgm/s

B) final velocity after collision = 2.87m/s

C) change in KE = -814.466J. 814.466J energy was lost.

Explanation:

Detailed explanation and calculation is shown in the image below

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