# Tag Info

2

Car collision "damage" usually goes with the energy in the zero momentum frame. In both cases that is (since in the zero momentum frame, the two cases are equivalent, assuming the masses of the cars are equal): $$E_1 = 2 \times \frac{1}{2} m v_{rel}^2 = m \left( 30 \frac{km}{h} \right)^2$$ Therefore a priori there is no difference between the two situations....

0

You are right in all you have written. But a more fruitful sentence could be: Friction always tries to prevent sliding. (Kinetic friction) If two boxes are sliding over one another, the friction on the top block will pull in the lower block's direction, and friction on the lower block will pull in the top block's direction. Friction tries to keep them ...

3

If a body is moving, this doesn't mean that a net force certainly must be exerted on it. It can move without any net force (First Law of Newton). You might say "How that body has started its motion without any net fore?" The answer is: "Equations of motion are moment equations, i.e. they are stated for moments not for a time interval ($F(t_0)=ma(t_0)$). So ...

2

You are correct in your definition of force. A car, not accelerating, has zero net force associated with it. However, if the car were to hit something--let's say it's me standing in the middle of the street--it would exert a net force on me, and by Newton's Third Law experience a net force equal in magnitude and opposite in direction. So how can an object ...

5

You are correct. The car has no net force on its environment, and the environment has no net force on the car. This is true of any object traveling with a constant velocity. This is even true in the vertical direction. There is a force of gravity pulling down on the car, and there is a force caused by the road pushing up on the car. If the car is not ...

4

It can't fall slower as the first cosmical speed (7.8 km/s), which is still very high. Although it would cause much smaller destruction as it would hit directly with the mean speed of the meteors (10-70km/s). The lower angle of the hit doesn't play a significant role, because considering its mass, the interaction with the atmosphere will be probably ...

0

You are orbiting at the surface of the earth, so $r$ is fixed. Since $r$ is fixed you are not free to change $a$; it is fixed by Newton's Law of gravitation $$a = \frac{F}{m} = G\frac{m_\mathrm{earth}}{r_\mathrm{earth}^2} = 9.8\;\;\mathrm{m/s}^2$$ Since $a$ is fixed and $r$ is fixed, $v$ is determined.

1

Assuming no air resistance, the only force acting on the satellite is the force of gravity, $\bf{F_g}$. The stated circular orbit also means that centripetal force, $\bf{F_c}$ is involved, which causes the satellite to follow the circular path. Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for ...

0

The answers are correct, the justification has some major problems. You have a ramp, so the slope is not 0. That means that the normal force (which is perpendicular to the ramp) and the gravity (vertical) are not collinear, so they cannot cancel each other. In a reference frame with an axis parallel to the ramp, the gravity has a component perpendicular to ...

0

I only watched the video to the point (about 15 min after the start) where Susskind he explained that according to Aristotle's "law of motion", the mass in a spring-and-mass system would move towards its equilibrium position with an exponentially decreasing velocity. Therefore, if you observe the mass when it is close to its equilibrium position and not ...

0

Assume all masses are the same and we have a perfectly symmetrical collision. It seems a logical assumption doesn't it? Then, try to understand the following conservation of energy equation. Use it to solve for $v_{af}$: $$\frac{1}{2}mv_{ai}^2 = \frac{1}{2}mv_{af}^2 + \frac{1}{2}mv_{b}^2 +\frac{1}{2}mv_{c}^2$$ Once you've done that, go back and solve the ...

0

You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A. In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common ...

1

The above isn't drawn correctly. And it is far too complicated in my opinion. I think I was able to solve this using high school physics: This bugged me a long time. I am surprised no one ever posted an answer on the internet. The length of his rope was 60.8 meters, assuming a drop above the target building of about 12 meters. The difference in height ...

1

There is extensive (100s of articles) lit. on the web regarding atmospheric drag and drag in general. A spherical bob and a very thin rod makes the problem very simple. One may use this graph: http://eis.bris.ac.uk/~memag/Teaching/Multi/dragcurve.pdf , and then calculated the Reynolds number to find the drag force using the formula given by the first ...

1

If we are dealing with the gravitational field of the earth, then because of the inverse square law, you should, in theory, need a tiny bit less force to raise an object from 50 m to 100 m, as you do lifting it from ground level to 50 m. But this difference in force is minute, I don't know offhand if we have instruments to measure it. But you can work it ...

0

In a rigid body, according to Goldstein's definition, the distance between any two constituent particles does not change. Work done is force times distance moved in the direction of the force. There is no relative movement in the direction of any force. Therefore, regardless of the form of the internal forces, no work is done by or against them.

3

All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn. Once again we start from the vis-viva equation: $$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1}$$ The parameter $a$ is the semi-major axis of the ellipse, and it is related to the ...

-1

Yes and I think this can be understood easily. When the same object is first lifted to 50m and then to 100 you can conclude that you need the same force but you need to apply it for a longer time(or distance). So, the force you need is same but there will be larger work done(consumption of energy) for larger distances

1

You have two questions. Here is the answer to your first question, "So why stone does not come back/reach to the centre of circle?" The answer is that the stone is accelerating toward the exact center of the circle. Velocity is composed of speed AND direction. As you noted, the velocity is always changing, however the speed is not. This is related to your ...

0

The problem was imperfectly posed, which leaves me trying to answer multiple issues; I'll try to answer all, but the order of answering is completely arbitrary. An ideal chain has inelastic links, but this would not affect a stretched coiled chain. Nevertheless, a stretched coiled chain would not give the answer you get from simple energy conservation ...

5

Recall the difference between the weak and strong Newton's third law, cf. e.g. this Phys.SE post. If the internal forces satisfy the weak Newton's third law (but not the strong Newton's third law, i.e. without the collinarity assumption), then it is not guaranteed that the internal forces do no work, cf. e.g. Fig. 1. ^ F | | 2 x-----------...

2

The motion (perpendicular to the wind) is called Ekman Transport. It results from a combination of the wind and the Coriolis Force (due to the rotation of the Earth), which together produce an underwater motion called the Ekman Spiral. Icebergs, which extend large distances below the surface, experience forces from the whole depth of the spiral. The ...

1

I thought about this in the molecule level, but then if molecules are at constant velocity then there is no "pushing" force that they apply on each other. To clarify my misunderstanding - I imagined the system at constant velocity at molecule level as if it was accelerating so molecules of A push molecules of B and also the opposite. If we are speaking about ...

1

If the blocks initial velocities are zero (i.e. the blocks start to move from rest), then it is impossible for block B to move with constant velocity. Because the only horizontal force acting on it (if there was) is friction force due to block A. We have: $$\Sigma F_B=m_Ba_B$$ If block B moves with constant velocity, then $a_B=0$ Thus, friction force acting ...

2

When you move the pen at constant velocity, you are correct that you apply $F=\mu_kmg$. This is true regardless of what constant velocity you choose. The difference is you have to supply more force initially to accelerate the pen to a faster speed, but once it is up to speed you only need $\mu_kmg$ to keep it at that speed. So it really isn't the force ...

0

If both the ropes are of identical material, then the fact that they have equal tension in them means, by symmetry, that they make equal angles with horizontal, call it $\theta$. Then the only equation you have is $mg=2Tsin\theta$. So unless $\theta$ is given you cannot solve the problem. If you make, say, $\theta$ small so that the hanging ropes are close ...

1

If you think of it in terms of conservation of momentum and collisions, the simplest version works just the same as tossing a handball at a on-coming freight train. The interaction is elastic, and the ball returns with the same speed it had going in in the center of momentum frame, but the center of momentum frame is moving in the ground frame, so the ball ...

1

For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for

1

You definitely don't need to use General Relativity to answer this question. It depends upon what you mean by "feel". If "feel" means "detectable by sophisticated instruments" then, yes, it can be "felt". But your body is not a very sophisticated detection instrument. According to what I've read elsewhere, the Earth speeds up by $1000$ $m/s$ as it moves ...

1

The systematic way to set up the equations is to draw a free body diagram for each mass. The FBD shows all the forces acting on that particular mass. You can then use Newton's second law to get the acceleration from the resultant force. The weight of mass 2 acts on mass 2, not on mass 1. Of course the weight of mass 2 will affect the motion of the system, ...

0

Yes, there is an error in your reasoning. The gravitational force on an object is always proportional to the mass of that object. So if you want the gravitational force on $m_1$, you use $m_1 g$. If you want the gravitational force on $m_2$, you use $m_2 g$. If you want the gravitational force on the combined system of $m_1$ and $m_2$, you use $(m_1 + m_2)g$,...

1

You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ...

1

There are two reasons. Inside the Sun, the force is no longer an inverse-square law. It actually grows linearly with $r$. The second reason is that Goldstein (as well as any other classical mechanics book) is interested in orbits with a non vanishing angular momentum with respect to the center of the Sun. An oscillation along a line passing through this ...

0

Your question is as follows: Why centripetal force does not increase the value of tangential velocity? Answer: Assume you have circular motion as in the case where a person, with her hand, twirls a ball on a string. The string connects her fingers to the ball as the ball travels in a circle around her hand. In this case, the force in the string is ...

9

Yes, it is possible to rise theoretically w.r.t a ground frame. But keep in mind that the rope-man system's center of mass must keep moving downwards because of the only external force acting on the system(gravity). The lighter M gets, the harder it will be for the man to rise, and it will become impossible in the limit the rope becomes massless. This is ...

0

What do people actually mean by "rolling without slipping"? This question and the answer should give you better insight into your question. Think about the definition of rolling without slipping given and how friction is created and works for rolling objects maybe make a free-body diagram.

5

Will a tennis ball go further if i hit it with the side of the racket? No. You want the racket to deform, not the ball. This means using the strings to elastically store energy and return it to the ball. The Ball The ball's deformation upon impact is undesirable because "a tennis ball is required by the rules of tennis to dissipate a fraction of ...

2

You are combinig two question, I am combining two answers. The system you describe consists of two points with masses. We know that: Every two points $A$ and $B$ lay on a single line; When line is rotated around axis intersecting it, all points of the line have same angular velocity $\omega$, except for the intersection with $\omega_0=0$; Centre of mass $... 0 I think you are comparing two pretty different cases. Motion of the Earth with respect to the Sun (and vise versa) is different from motion of a block on a table with respect to it. In the former, both of the Earth and Sun experience rotational motion and cannot be assumed as a particle without approximation error. But in the later, the block doesn't ... 0 To impart certain amount of kinetic energy to tennis ball you will have to do work on it. Now, work done= Power x Time. The difference between hitting using netting and hitting sideways is that, in the former case you have more contact time with the tennis ball available, so you need smaller average power to do necessary amount of work. When you hit sideways,... -1 You have two unknowns$A$and$\phi_o$so you need two equations. You are given the displacement$y$at$t=0$which gives you one of the equations. If you differentiate the equation for the displacement$y$with respect to time$t$this will give you an expression for the velocity which is you second equation. 2 I think the OP has made a mistake in applying second law of Newton. The law (about a particle) says: $$\Sigma \vec F=m\vec a$$ As it is seen, this is a vector equation. This means that corresponding components of both side of the equation must be equal. Although it is not said in the law's body, but it is obvious that we must write the equation above with ... 1 A properly drawn free-body diagram will have a tension force vector acting along the line of the rope, toward the pivot point and a gravity vector acting straight down. If you establish a coordinate system which is instantaneously parallel and perpendicular to the rope, you then will decompose the gravity vector into two components ($mg$times trig ... 6 Here's some tennis racket physics from Rod Cross, including links to several Am. J. Phys articles (the physics educators' journal, thus excellent for learning from) and this excellent diagram: There are at least three "sweet spots": The node, at the center of the strings, is a point where the natural standing waves in a vibrating racket don't have any ... 3 The rope pulls just enough that the pendulum doesn't fall to the ground, but follows an arc. The following picture shows you how work out the force for the static case (no motion of the pendulum): However, you need to take account of the fact that the pendulum is moving in an arc. When something moves in an arc, you need an additional force$F=\frac{mv^...

2

Which way does the y axis point? If the y axis is chosen to point up, (having the positive direction upwards) then you are right, normal force should be positive (it points upwards as well) and weight negative. Is it chosen to point down, then normal force is negative (points in the negative direction along the axis) and weight positive. Remember that ...

3

Centripetal force is the name of the force that points towards the centre. This is in the radial direction. Tangential velocity is, as the name suggests, a velocity direction tangent to the circle. The radial and tangential directions are by definition always perpendicular - in the same way that the x and y axes are. You are of course right that if any ...

2

The centripetal acceleration always points toward the center of the circle. In this case, the center of the circle is below the car, so the centripetal acceleration points downward. Now, you'll notice that, in the given solution, the centripetal acceleration term is positive. That means the writer has chosen a coordinate system where positive is downward, ...

7

Kepler's third law is irrelevant here. It applies to many (small) planets orbiting a (large) central star, not to a binary star system. If the stars have masses $m_1$ and $m_2$ and the radii from the COM are $r_1$ and $r_2$, then from the definition of the COM, $m_1r_1 = m_2r_2$. The (gravitational) central force $F$ acting on each star is the same, but ...

-3

Interesting indeed... My guess is if you hit the ball on the frame it should go farther. Here's my thinking: Hitting with the frame causes all the compression energy to go into the ball. The frame does not compress at all. Hitting on the stringing causes both the ball and strings to compress, which expends more energy than compressing the ball alone. ...

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