A bullet of mass 12 g moved at a speed of 450 m/s - Edexcel - A-Level Physics - Question 15 - 2023 - Paper 1

To find the maximum change in vertical height, we can use the principle of conservation of linear momentum and energies:

1. Using conservation of momentum, the initial momentum of the bullet must equal the combined momentum of the bullet and block after the collision:

   $0.012 ext{ kg} imes 450 ext{ m/s} = (0.012 ext{ kg} + 2.5 ext{ kg}) v_f$

   Rearranging gives:

   v_f = rac{0.012 imes 450}{0.012 + 2.5} = 0.012 ext{ kg}/(2.512 ext{ kg}) imes 450 ext{ m/s}

   $v_f ext{ approx } = 0.021 ext{ m/s}$

2. The maximum height ( h) can be determined using energy conservation (kinetic energy to potential energy):

   rac{1}{2} mv^2 = mgh

   Rearranging gives:

   h = rac{v^2}{2g}

   where g = 9.81 m/s².

   Substituting the values gives:

   h = rac{(0.021)^2}{2 imes 9.81} ext{ m} ext{ approx } = 0.000022 m

   Therefore, the maximum change in vertical height of the wooden block is approximately 0.000022 m.

The explanation involves several considerations:

1. The final momentum/velocity of the bullet is negative after hitting the steel block, meaning it has a greater change in momentum compared to the wooden block because.
2. When the bullet hits the steel block, the change in momentum is greater because it rebounds back, unlike in the wooden block case where the bullet was embedded.
3. Since the steel block is harder, it experiences greater force during the impact, thus it gains greater initial velocity after the bullet rebounds.
4. This leads to greater kinetic energy being converted into potential energy, resulting in the steel block moving through a greater maximum vertical height than the wooden block.

Ultimately, the greater post-collision speed of the steel block leads to more potential energy at its apex height.