# F As a rocket accelerates it experiences a force F which is responsible for the acceleration. The reaction force to this force F (the other force in the.

## Presentation on theme: "F As a rocket accelerates it experiences a force F which is responsible for the acceleration. The reaction force to this force F (the other force in the."— Presentation transcript:

F As a rocket accelerates it experiences a force F which is responsible for the acceleration. The reaction force to this force F (the other force in the force pair) acts on the rocket exhaust, the gases ejected out the rear of the rocket. If the acceleration of the rocket at a given time is A and its mass is M, while the gas ejected over the same time has mass m and acceleration a, which of the following is true? a)A = (m/M) a b) A = - (M/m) a c) A = - (m/M) a d) A = (M/m) a Rocket -F-F

Correct Answer: C Since the two forces in the force pair are equal and opposite, it follows from Newton’s second law (F = m a) that m a = - M A, which gives us A = - (m/M) a In most cases the thing that you push off has more mass and thus doesn’t accelerate as much as you do (like when I jump off the floor, or when the book and the Earth pull on each other). But in a rocket the gas is much lighter than the rocket, so its acceleration is actually much greater. However the gas is so light that air resistance quickly slows it down again so it doesn’t travel very far. Another unusual thing about rocket motion is that the mass of the gas ejected means that the rocket loses mass as it accelerates. Usually we write Newton’s second law as F = m a = m  v/  t, where  v is the change in the rocket’s velocity in the time  t. But this assumes the rocket’s mass remains constant, which it doesn’t. So a better way to write the equation is F =  (m v)/  t =  p/  t where p is called the momentum of a body, where p = m v. So what a force does is change the momentum of a body.

m1=10 kg m2=5 kgF Box 1 Box 2 F2F1 Suppose you are pushing two boxes across the floor. Let’s suppose you push with a force F = 20 newtons. Obviously you are only applying a force to Box 1. Since Box 2 must move also, the force on it must come from Box 1 (let’s call that force F2). You don’t get that help for free however, because Box 2 pushes back on box 1 with a force we’ll call F1. Which of the following is a correct equation for the acceleration of box 1, a1? a)a1 = F/(m1+m2) b)a1 = F2/m2 c)a1 = (F + F1)/m1 d)All of the above

Correct Answer: D Why are all of these equations correct? a)a1 = F/(m1+m2) – In this case we treat the two boxes as a single object. When applying Newton’s second law to one object of connected group of objects, ignore all INTERNAL forces. Only EXTERNAL forces can accelerate the whole object. So only F is involved here, and the mass of the two boxes together is m1 + m2. b)a1 = F2/m2 – The two boxes must accelerate at the same rate, so a1 = a2. The acceleration of box 2 is easy to calculate, because the only force on box 2 is F2 and its mass is m2. c)a1 = (F + F1)/m1 – Looking at box 1 by itself we see that there are two forces acting on it, F and F1, which oppose each other. The net force on box 1 is the sum of these two vectors. Since they oppose each other, the magnitude of a1 = (F – F1)/m1. m1=10 kg F Box 1 Box 2 F2F1

Now we can calculate the acceleration of the boxes, from the equation a1 = F/(m1+m2) We know that F = 20 newtons, m1 = 10 kg and m2 = 5 kg, which means That a)a1 = 2 m/s/s b)a1 = 1.333 m/s/s c)a1 =.667 m/s d)a1 = 1 m/s/s

Correct answer: B Note that it couldn’t be a whole number, because 15 doesn’t go into 20. Also the units for answer d are wrong (m/s is a speed or velocity).

If we want to calculate the forces between the two boxes F1 and F2. What should our strategy be? a)We need to use the equations we derived in the previous question to set a1 = F/(m1+m2) = F2/m2 b)We need to use Newton’s third law to show that F1 = -F2 c)We need to use the fact that the mass of the small box (m2) is only 1/3 the total mass of the two boxes (m1+m2). d)Actually, all of the above again.

Correct Answer: D again! But A is the main point, the other two being fairly obvious. Clearly we can use the equation given in a) to calculate F2 and get a1 from this. Strictly speaking we are really making use of Newton’s third law (b) in the way we set up the whole problem. The third point c) is, in this case, another way of saying that we know the masses of the boxes. But note that we don’t really need to know the total masses to get the two forces, all we need to know is the relative sizes of the two boxes.

Now let’s suppose we decided we needed to push the boxes back the other way. We go around to the other side and start pushing on the smaller box to push the whole box across the room. If we push with the same force as before, m1=10 kg F Box 1 Box 2 F2F1 What is the acceleration of the two boxes this time? a)Half what it was in the other case (because Box 2 is half the mass of Box 1) b)Twice what it was in the other case (because Box 1 is twice the mass of Box 2) c)The same as in the other case (because the total mass is the same) d)All of the above

Correct answer: C As far as the acceleration of the two boxes goes it is only the external force and the total masses of the two boxes which matters, because of the equation a1 = F/(m1+m2)

Suppose I get fed up of pushing boxes across the floor and decided to attach a rope to a single box and pull it across the floor. Rope F W N  Because two of the forces acting on the box (the weight W and the normal force N) are vertical, it makes sense to use your usual x and y coordinates. We can easily figure out the components of F, the pulling force which acts along the rope, as long as we know the angle of the rope to the horizontal. Obviously the x component of F accelerates the box. In typical cases, what does the vertical component of F do? a)Lessens the weight W of the box b)Lessens the size of the normal force N c) Changes the angle of N d) Has no effect

Correct answer: B You can’t change the weight of a block just be pulling on it, but you do change the force pressing into the floor, which changes the force of the floor pressing back.

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