The force on a bullet along the barrel of a firearm is given by the formula F = [740 - (2.3 x 10 5 s -1 )t] N over the time interval t = 0 to t = 3.0 x 10 -3 s. What is the recoil speed of the 4.5-kg gun?

Hey, everyone in this problem, a cannon shot a cannonball, the force applied on the cannonball over the length of the barrel can be modeled by F is equal to 890 minus 5.7 multiplied by 10 to the exponent five inverse seconds multiplied by time T that's measured in newtons. The model is only valid for T ranging from 0 to 2 multiplied by 10 to the exponent negative three seconds. As a result of this impulse given to the cannonball, it comes out of the barrel of the cannon with a speed of 120 m per second. The mass of the cannonball is 3 kg. The mass of the cannon is 5 kg. And we're asked to calculate the speed with which the cannon recoiled after firing. We're given four answer choices all in meters per second. Option A 0.08 option B 0.72 option C 1.4 and option D 13. Now, in this case, there's a couple of different ways to approach this problem. We could look at the impulse and do some calculations around the impulse and its relationship to the momentum something simpler we can do is just look at the conservation of momentum. And we're gonna assume that we have no external forces acting here in which case, momentum will be conserved. And if momentum is conserved, we know that the initial momentum P not is just gonna be equal to the final momentum. Pf now the momentum we have acting is gonna consist of two parts and we have the momentum of the cannonball and we have the momentum of the cannon itself. So we have P not B plus P, not C. OK. We're gonna use subscript B to indicate canon ball and we'll use subscript C to indicate just the canon. OK. So that's our initial momentum that's gonna equal our final momentum PFB plus PF. Now let's recall that the momentum is given by mass multiplied by velocity. So we can use that in each of our terms. We get MB V not be was MC V Nazi. That's gonna be equal to M BVFC plus MC, oops VF B, sorry plus M CV FC. OK. So let's think about all of these velocities that we have in this equation. Initially, the cannonball is not moving, it's inside of that cannon until that impulse is experienced or exerted on that cannonball, it is not moving. So the initial velocity of the cannonball is zero. And same with thing with the cannon. The cannon is also not moving. So our initial momentum on the left hand side is just gonna be zero. Ok. So we have zero on the right hand side, we have the mass of the cannonball, which we're told is 3 kg multiplied by its final velocity. We know that it has a speed of 120 m per second. We're gonna take that to be a positive velocity, ok? So the direction of the cannonball is positive and we're gonna add the mass of the cannon itself 500 kg multiplied by its velocity VFC. And that is the value we are looking for. Now, we're gonna move our 500 kg VFC to the other side and we are gonna simplify the left hand side. We get 500 kg multiplied by VFC is going to be equal to negative 360 kg meters per second. The last thing we're gonna do divide by 500 kg and the velocity of the cannon after it fires this cannonball is going to be negative 0.72 meters per second. Now, this is a velocity and we have this negative sign and all the negative sign indicates is that this is moving in the opposite direction as the cannonball, ok? We've assumed that the cannonball has a positive velocity. That means that this negative velocity is the opposite direction and that makes sense. Ok? The cannon is recoiling. So it's moving in the opposite direction we're moving backwards. Ok? Now, the question is not asking for the velocity it's asking for the speed. So the last thing we need to do is just to take the magnitude. So our speed VFC just gonna be the magnitude of this velocity which is gonna be 0.72 m per second. Taking a look at our answer choices. We can see that the correct answer is going to be option B 0.72 m per second. Thanks everyone for watching. I hope this video see you in the next one.

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11. Momentum & Impulse

Push-Away Problems

Physics

11. Momentum & Impulse

Push-Away Problems

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Problem 24e

Textbook Question

The force on a bullet along the barrel of a firearm is given by the formula F = [740 - (2.3 x 10⁵ s^-1 )t] N over the time interval t = 0 to t = 3.0 x 10^-3 s. What is the recoil speed of the 4.5-kg gun?

Verified step by step guidance

Understand the problem: The force acting on the bullet is given as a function of time, F(t) = 740 - (2.3 × 10⁵ s⁻¹) t. The goal is to find the recoil speed of the gun, which can be determined using the principle of conservation of momentum. The mass of the gun is 4.5 kg, and the time interval is t = 0 to t = 3.0 × 10⁻³ s.

Step 1: Calculate the impulse on the bullet. Impulse is the integral of force over time, which is given by: I = ∫ F(t) dt from t = 0 to t = 3.0 × 10⁻³ s. Substitute F(t) = 740 - (2.3 × 10⁵ s⁻¹) t into the integral: I = ∫ [740 - (2.3 × 10⁵ s⁻¹) t] dt.

Step 2: Solve the integral. Break it into two parts: I = ∫ 740 dt - ∫ (2.3 × 10⁵ s⁻¹) t dt. The first term integrates to 740t, and the second term integrates to (2.3 × 10⁵ s⁻¹) t² / 2. Evaluate these expressions at the limits t = 0 and t = 3.0 × 10⁻³ s.

Step 3: Use the impulse-momentum theorem. The impulse on the bullet is equal to the change in momentum of the bullet. Since the gun and bullet are part of the same system, the momentum of the gun will be equal and opposite to the momentum of the bullet. Let the recoil speed of the gun be v_g. The momentum of the gun is p_g = m_g v_g, where m_g = 4.5 kg.

Step 4: Relate the impulse to the gun's recoil speed. From conservation of momentum, the impulse on the bullet is equal to the momentum of the gun: I = m_g v_g. Solve for v_g: v_g = I / m_g. Substitute the value of I (calculated from Step 2) and m_g = 4.5 kg to find the recoil speed of the gun.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Third Law of Motion

Newton's Third Law states that for every action, there is an equal and opposite reaction. In the context of a firearm, when the bullet is propelled forward by the force exerted by the expanding gases, the gun experiences an equal force in the opposite direction, resulting in recoil. Understanding this principle is crucial for calculating the recoil speed of the gun.

Impulse and Momentum

Impulse is the change in momentum of an object when a force is applied over time. The impulse experienced by the bullet as it travels down the barrel can be calculated by integrating the force over the time interval. This impulse is equal to the momentum gained by the bullet, and by conservation of momentum, it can be used to determine the recoil speed of the gun.

Kinematics of Recoil

The kinematics of recoil involves analyzing the motion of the gun after the bullet is fired. The recoil speed can be calculated using the relationship between the mass of the gun and the momentum of the bullet. By applying the conservation of momentum principle, the recoil speed can be derived from the bullet's mass and velocity, providing insight into the gun's motion post-firing.

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The force on a bullet along the barrel of a firearm is given by the formula F = [740 - (2.3 x 105 s-1 )t] N over the time interval t = 0 to t = 3.0 x 10-3 s. What is the recoil speed of the 4.5-kg gun?

Key Concepts

Newton's Third Law of Motion

Impulse and Momentum

Kinematics of Recoil

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The force on a bullet along the barrel of a firearm is given by the formula F = [740 - (2.3 x 10⁵ s^-1 )t] N over the time interval t = 0 to t = 3.0 x 10^-3 s. What is the recoil speed of the 4.5-kg gun?