A10 - g bullet moving directly upward at 1000 m/s strikes andpasses
through the center of mass of a 5-kg block initially at rest. thebullet
emerges from the block, moving directly upward at 400 m/s.

a) what is the speed of the box immediately after the collisionwith the
bullet?
b) to what maximum height does the block then rise above itsinitial
position?
c) how much energy is lost during the collision?

a)  v2f = 1.2 m/s, b)   h = 7.35 10⁻² m and c)  ΔK = -4196.4 J ,

Explanation:

a) This problem must be solved with the conservation of the moment. Let's define the system as the one formed by the bullet plus the block, in this system all the forces are internal therefore the moment conserves, let's write the moment in two moments before and after the crash. In general these shocks are very fast, so it can be assumed that the box does not move during the crash.

The data they give us is the mass of the bullet (m = 0.010 kg), the initial and final velocities of the bullet (v1o = 1000 m / s and v1f = 400 m / s) and the block gives us the mass M = 5 kg and its initial velocity v2o = 0 m / s

Before the crash

      po = m v₁₀

After the crash

     pf = m + M

     p₀ = pf

     m v₁₀ = m + M

     v2f = m (v₁₀ - ) / M

     v2f = 0.010 (1000-400) / 5

     v2f = 1.2 m / s

b) Having the speed of the block we can use the law of conservation of energy to find the height. Let's write the mechanical energy of the block just after the crash and at the point of maximum height

Initial. Just after the crash

      v =

      Em1 = K = ½ M v²

Final. At maximum height

     Em₂ = U = M g h

     Em₁ = Em₂

     ½ M v² = M g h

     h = ½ v² / g

     h = ½ 1.2² / 9.8

     h = 0.0735 m

     h = 7.35 10⁻² m

c) Let's calculate the kinetic energy before and after the crash

Before

     K₀ = ½ m v₁₀²

     K₀ = ½ 0.01 1000²

     K₀ = 5000 J

Final

    = ½ m v1f² + ½ M v2f²

    = ½ 0.010 400² + ½ 5 1.2²

    = 800 + 3.6

  = 803.6 J

We can give the amount of energy that is lost as the subtraction of the two energies or as the fraction of lost engoa

    ΔK = -K₀

    ΔK = 803.6 - 5000

    ΔK = -4196.4 J

    / K₀ = 803.6 / 5000

    / K₀ = 0.16