Abullet with mass m = 0.1 kg grams hits a ballistic pendulum with length l = 3 meters and mass m = 2 kg and lodges in it. when the bullet hits the pendulum it swings up from the equilibrium position and reaches an angle 15 degrees at its maximum. in this problem, you will determine the bullet’s velocity. first, i want to ask you a couple of questions:

while the bullet is embedding into the block, is energy conserved? why or why not?
b. while the bullet is embedding into the block, is momentum conserved? why or why not?

c. when the pendulum swings up to its maximum angle, is energy conserved? why or why not?

d. when the pendulum swings up to its maximum angle, is momentum conserved? why or why not?

e. find the velocity of the bullet.

e)     v₁ = 29.7 m / s

Explanation:

Let's propose the solution of the problem, let's start at the moment

Initial

        p₀ = m v₁

Final

      = (m + M) v

The moment is preserved

      p₀ =  

      m v₁ = (m + M) v

      v = m / (m + M) v₁        (1)

a) energy is conserved

Let's look for kinetic energy

Initial

      K₀ = ½ m v₁²

Final

      = ½ (m + M) v²

Let's replace v

      = ½ (m + M) [m / (m + M) v₁]²

      = ½ m² / (m + M) v₁²

Let's look for the relationship of these energies

     Ko / = ½ m v₁² / (½ m² / (m + M) v₁²)

     Ko / = (m + M) / m = (1 + M / m)²

The kinetic energy changes therefore it is not conserved in the process, the missing energy is converted into potential heat energy during the crash

b) The impulse is conserved because the system is defined as formed by the two bodies and the externals are of action and reaction, so for the complete system the sum is zero and the moment does not change in value

c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential

d) when the pendulum oscillates the speed changes from v to zero, so the moment is not conserved, this is because there is an external force acting on the system, the force of gravity

e) For this part let's start at the end of the movement

It is system (bullet + block) moves, energy is conserved

Final. Highest point

          = U = (m + M) g h

Initial. Lowest point

          Em₀ = K = ½ (m + M) v2

          Em₀ =  

          ½ (m + M) v² = (m + M) g h

          v = √ 2gh

Let's look for the height (h) by trigonometry

         Cos 15 = x / L

          h = L-x

          h = L - L cos 15

          h = L (1- cos 15)

We replace

          v = √ (2gL (1- cos 15))

Now we use equation (1) of momentum conservation

         v = m / (m + M) v1

         v₁ = (m + M) / m v

         v1 = (0.1 +2.0) /0.1 RA (2 9.8 3 (1- cos 15))

         v₁ = 21 √ (2.00)

        v₁ = 29.7 m / s

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