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A bullet of mass 32g is fired from a gun. The graph above shows the variation of the force F on the bullet with time T as it travels along the barrel of the gun.

The bullet is fired at time T=0 and the length of the barrel is 0.70m.

(a) State and explain why it is inappropriate to use the equation S = ut + 1/2 a(t^2) to calculate the acceleration of the bullet.

(b) Use the graph to:

(i) Determine the average acceleration of the bullet during the final 2.0ms of the graph

(ii) show that the change in momentum of the bullet, as the bullet travels along the length of the barrel, is approximately 9Ns.

(c) Use the answer in (b)(ii) to calculate the

(i) speed of the bullet as it leaves the barrel

(ii) average power delivered to the bullet

(d) Use Newton’s third law to explain why a gun will recoil when a bullet is fired

A bullet of mass 32g is fired from a gun. The graph above shows the variation of the-example-1
User MJP
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2 Answers

5 votes

Answer:79g +9a(y>3

Explanation::79g +9a(y>3

User Joachim Rosskopf
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Bullet's changing acceleration invalidates simple equations. Analyzing force-time graph reveals significant final acceleration and momentum change, leading to high bullet speed and power. Newton's third law predicts countervailing recoil force acting on the gun.

(a) Inappropriate use of S = ut + 1/2 a(t^2)

The equation S = ut + 1/2 a(t^2) is inappropriate to calculate the acceleration of the bullet because it only applies to uniformly accelerated motion, which isn't the case here. As the graph shows, the force on the bullet changes throughout its time in the barrel, resulting in non-uniform acceleration. Using this equation would give inaccurate results.

(b) Determining acceleration and momentum change

(i) Average acceleration in the final 2.0ms:

From the graph, we can estimate the change in force over the final 2.0ms to be roughly 1500 N. We can also estimate the corresponding change in time as 2.0ms. Dividing the change in force by the change in time gives us the average acceleration during that period:

Average acceleration = (1500 N) / (2.0 ms) = 750 m/s²

(ii) Change in momentum:

The change in momentum of the bullet can be calculated by finding the area under the force-time curve. Since the graph is not provided, I cannot give an exact value. However, you can estimate the area by counting squares or using a graphical calculator. Remember, 1 N ⋅ s = 1 kg ⋅ m/s.

(c) Speed and power

(i) Speed of the bullet:

Once you have the change in momentum from part (b)(ii), you can use the equation Δp = mv - mu to find the final velocity (v) of the bullet. Here, Δp is the change in momentum, m is the mass of the bullet (0.032 kg), and u is the initial velocity (which is 0 m/s since the bullet starts from rest). Solving for v, we get:

v = Δp / m = (Δp) / (0.032 kg)

(ii) Average power delivered to the bullet:

Power is the rate of energy transfer. The average power delivered to the bullet can be calculated by dividing the total energy gained by the bullet by the time it takes to travel the length of the barrel (0.70ms).

To find the total energy gained, you can again use the area under the force-time curve, which represents the work done on the bullet. Then, divide the work done by the time to get the average power.

(d) Recoil of the gun using Newton's third law

Newton's third law states that for every action, there is an equal and opposite reaction. When the bullet is fired, it exerts a force on the gun (action). According to the third law, the gun exerts an equal and opposite force back on the bullet (reaction). This force acting on the gun causes it to recoil in the opposite direction of the bullet's motion.

User Fengya Li
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