Question

1. Is the collision between the ball and the pendulum elastic or inelastic? Justify your answer by calculating the kinetic energy of the system before collision using the value of vxo in the experiment and the kinetic energy just after collision using the experimental value of h in Eq. 9.2.

Answer 1

Answer:

So energy is not conserved and inelastic shock

Explanation:

In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.

Let's find the kinetic energy just before the crash

         K₀ = ½ m vₓ₀²

After the crash we can use the law of conservation of energy

Starting point. Right after the crash, before starting to climb

        Em₀ = K = ½ (m + M) v₂²

Final point. At the maximum height of the pendulum

       $Em_{f}$ = U = (m + M) g h

Where m is the mass of the bullet and M is the mass of the pendulum

        Em₀ = Em_{f}

        ½ (m + M) v₂² = (m + M) g h

        v₂ = √ 2g h

Now we can calculate the final kinetic energy

       K_{f} = ½ (m + M) v₂²²

       K_{f} = ½ (m + M) (2gh)

The relationship between these two kinetic energies is

      K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

      K₀ / K_{f} = m / (m + M)  vₓ₀² / 2 g h

We can see that in this relationship the Ko> Kf

So energy is not conserved and inelastic shock

Answer 2

Here the energy is not conserved and should be considered as the inelastic shock.

Reason of energy should not be conserved:

Here

We assume to determine the kinetic energy just prior to the crash

K₀ = ½ m vₓ₀²

After the crash, we use the law of conservation of energy

Starting point. Right after the crash, prior to starting to climb

Em₀ = K = ½ (m + M) v₂²

Now

Final point. At the maximum height of the pendulum

So,

= U = (m + M) g h

here

m is the mass of the bullet

and M is the mass of the pendulum

So,

       Em₀ = Em_{f}

       ½ (m + M) v₂² = (m + M) g h

       v₂ = √ 2g h

Now the final kinetic energy

      K_{f} = ½ (m + M) v₂²²

      K_{f} = ½ (m + M) (2gh)

So,

The relationship between these two kinetic energies is

     K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

     K₀ / K_{f} = m / (m + M)  vₓ₀² / 2 g h

here the relationship is the Ko> Kf

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