3.1 Two bullets are fired, one from a pistol and the other from a rifle, as shown in the diagram below - NSC Technical Sciences - Question 3 - 2022 - Paper 1

The bullet fired FROM THE PISTOL will have the greater change in momentum. This is because the pistol has a shorter barrel, resulting in a shorter time during which the force acts on the bullet, leading to a greater change in momentum compared to the longer barrel of the rifle.

The principle of conservation of linear momentum states that the total momentum of a closed system remains constant if no external forces act upon it. In other words, the sum of the momenta before an interaction is equal to the sum of the momenta after the interaction.

Using the conservation of momentum:

Before: $(m_{rifle} + m_{bullet})v_{initial} = 0$ After: $(m_{rifle} v_{rifle} + m_{bullet} v_{bullet}) = 0$ Substituting, we have: $(1,2 v_{rifle}) + (0,03 * 330) = 0$ Solving for $v_{rifle}$:

$v_{rifle} = -\frac{(0,03)(330)}{1,2} \\ v_{rifle} = -8,25 m·s^{-1}$ This indicates that the rifle recoils in the opposite direction.

An elastic collision is one in which both kinetic energy and momentum are conserved, whereas in an inelastic collision, only momentum is conserved, and kinetic energy is not conserved. In elastic collisions, the objects rebound off one another without any loss of kinetic energy, while inelastic collisions often result in some kinetic energy being transformed into other forms of energy, such as heat or sound.

Using the formula for force:

$F_{net} = \frac{\Delta p}{\Delta t}$ With the change in momentum given by: $\Delta p = mv_f - mv_i$ Substituting the values: $m = 70 kg, \Delta v = 0 - (-12.5) = 12.5 m·s^{-1}, \Delta t = 0.03 s$ Calculate the change in momentum: $\Delta p = 70 kg * 12.5 m·s^{-1} = 875 kg·m·s^{-1}$ Now substitute into the force equation: $F_{net} = \frac{875}{0.03} = 29166.67 N$ Thus, the force exerted onto his legs by the Earth is approximately 29166.67 N upwards.