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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

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3.1 Two bullets are fired, one from a pistol and the other from a rifle, as shown in the diagram below. Assume that the pistol and the rifle exert the same force on ... show full transcript

Worked Solution & Example Answer: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

Step 1

3.1.1 State Newton's Third Law in words.

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Answer

Newton's Third Law states that for every action, there is an equal and opposite reaction. This means that if object A exerts a force on object B, then object B exerts a force of equal magnitude and opposite direction back on object A.

Step 2

3.1.2 When the shot is fired, which bullet will have the greater change in momentum? Write down only FROM THE PISTOL or FROM THE RIFLE. Explain the answer.

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From the Pistol. The bullet fired from the pistol will experience a greater change in momentum because it is propelled over a shorter period of time compared to the bullet fired from the rifle, which has a longer barrel. Consequently, the shorter time frame results in a higher acceleration and, thus, a greater change in momentum.

Step 3

3.2.1 State the principle of conservation of linear momentum in words.

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Answer

The principle of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant if no external forces act on it. Therefore, the momentum before an event, such as a collision or gunshot, is equal to the momentum after.

Step 4

3.2.2 Calculate the velocity at which the rifle recoils (kicks back) when the shot is fired.

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Using the principle of conservation of momentum:

extbefore:mrimesvr+mbimes0=mrimesvr+mbimesvb ext{before: } m_{r} imes v_{r} + m_{b} imes 0 = m_{r} imes v_{r} + m_{b} imes v_{b}

Substituting in the values: (1.2+0.03)imes0=1.2imesvr+0.03imes330(1.2 + 0.03) imes 0 = 1.2 imes v_{r} + 0.03 imes 330

This simplifies to: 0=1.2imesvr+9.90 = 1.2 imes v_{r} + 9.9

Solving for the recoil velocity of the rifle, we find:

ightarrow v_{r} = -8.25 ext{ m/s}$$ (This indicates that the rifle recoils backward.)

Step 5

3.2.3 Distinguish between the concepts elastic and inelastic collisions.

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Answer

In elastic collisions, both momentum and kinetic energy are conserved, meaning that the objects bounce off each other without any loss of total energy. In inelastic collisions, momentum is conserved, but kinetic energy is not; some of the energy is transformed into other forms (such as heat or deformation), and the objects may stick together after the collision.

Step 6

3.3 Calculate the force exerted onto his legs by the Earth.

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Answer

We can calculate the force exerted on the firefighter's legs using Newton's second law. First, we need to determine the change in momentum as he comes to a stop:

Initial velocity, vi=12.5extm/sv_i = 12.5 ext{ m/s} Final velocity, vf=0extm/sv_f = 0 ext{ m/s} Time, t=0.3extst = 0.3 ext{ s}

The change in momentum, extImpulse=mimesextchangeinvelocity=70imes(012.5)=875extkgm/s ext{Impulse} = m imes ext{change in velocity} = 70 imes (0 - 12.5) = -875 ext{ kg m/s}

The average force can be calculated as:

ightarrow F_{net} = -2916.67 ext{ N}$$ The negative sign indicates direction, while the magnitude of force exerted onto his legs by the Earth is approximately $2916.67 ext{ N}$ acting upwards.

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