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Ok I get it, if the spaceship does not attract the package, its what we get (sorry for my rusty algebra :-)) But Gneil has a point, if the package is acted by the planet's attraction, there is energy conservation and we should have: ##\frac{1}{2}mv^2 - \frac{GmM}{R} = \frac{1}{2}mv_0^2 -...

Yes, plus I made a mistake. When the package is at the surface of the planet, the spaceship attracts the package with the force : ##\vec F_s = - GmM_s \frac{5R\hat x + R\hat r}{|5R\hat x + R\hat r|^3} ## When the package just arrives at the surface of the planet ##\hat x## and ##\hat r## are...

Also, we want no radial acceleration for the package. With Newton second law: ## m\frac{v^2}{R} = \frac{Gm}{R^2} (M +\frac{ M_{\text{s}}}{26}) \Rightarrow v = \frac{\sqrt{G}}{R} \sqrt{(M +\frac{ M_{\text{s}}}{26})} ## Is that correct ?

Hmm, am I wrong here ? Ok, thanks!

Homework Statement A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at distance 5R from the center of the planet, the ship fires an instrument package with speed v0 (at angle ##\theta## with the line passing through the spaceship and the...

Thank you very much for your help ! I have this strange feeling that I got mentally blocked, it reminds me a ski tow problem :-)

Ok, with your explanations and those from TSny (stool example), maybe I see clearer. a)The sand into drum A exerts a force on it, the centrifugal force TSny mentioned, radially. Drum A would rotate freely without this applied force, so the external torque of Drum A is 0. Then, the angular...

I did symetrically with drum B, so that I get and expression of La and Lb. Then, the argument is that there is conservation of angular momentum for both La and Lb. From the point of view of "drum A + sand" subsystem, the only exterior forces acting on it are the weight of the sand rings into...

Thanks ! But why is it wrong? "Drum A + sand" sub-system includes drum A, the ring of sand that is into drum A, and the ring of sand that is into drum B. So the angular momentum of this sub-system about the axis of rotation is : ## L_a = L_{\text{drum A}} + L_{\text{Sand in A}} + L_{\text{Sand...

I've made an attempt at a solution, which is not correct. Can you explain where it starts to be wrong and why ?

I don't understand how it relates to the problem, what did you want to tell me with the stool example ?

If my moment of inertia does not change and angular momentum is conserved, then angular velocity does not change.

When I am rotating with the weights in my hand they exert a downward force on me. That force is parallel to the axis of rotation so there will be no torque about this axis. When I release the weight, the moment of inertia does not change if I am the system (lol, it sounds crazy)

Yes it is a typo, sorry :) In this situation, I think that angular momentum is conserved because there are no external forces in the plane of motion. If I drop the masses, my moment of inertia will decrease, so the angular velocity must increase in order to satisfy conservation of angular...

Homework Statement A drum of mass MA and radius a rotates freely with initial angular velocity ωA(0). A second drum with mass MB and radius b > a is mounted on the same axis and is at rest, although it is free to rotate. A thin layer of sand with mass Ms is distributed on the inner surface of...

Ahhh, ok ! I understand now. Thank you !

Thanks ! I've just checked on the internet, it's the same thing with 1/r factored out of the determinant

Thank you for your counter example ! ( Last post is too advanced for me, sorry, I wish I could reflect on that )

Hello, If there is no mistakes, I find that the work on this path is ##W = -\pi \vec F_0.\vec x_0 \neq 0##, which proves that the force is non conservative. Thanks, that makes it clear !

The force depends on time and not on position, so the curl test is useless. Remains the integral test: ##\oint \vec F.d\vec r = \int_0^T \vec F.\vec v\ dt = \int_0^T \vec F_0.\vec v \sin(at)\ dt## But I don't know what to do with that. I found something, with the assumption that F is a net...

Homework Statement a - ##\vec F = \vec F_0 \sin(at) ## b - ##F = A\theta \hat r##, A constant and ## 0 \le \theta < 2\pi ##. ##\vec F## is limited to the (x,y) plane c - A force which depends on the velociity of a particle but which is always perpendicular to the velocity Homework Equations...

I find 3.13m/s for v

up

Homework Statement A particle of mass m moves in a horizontal plane along the parabola ##y = x^2##. At t=0, it is at the point (1,1) with speed v0. Aside from the force of constraint holding it to the path, it is acted upon by the following external forces: A radial force: ##\vec F_a = -A...

There is a typo here, forgot a factor 11/20, but luckily it does not interfere with the rest. Got (b) too thanks to you ;-) By conservation of momentum in laboratory coordinates, if the neutron is ejected forward, then the helium nucleus and the neutron follow the same direction, which...

Thanks ! Got it for part (a) ! The center of mass has constant velocity vector ##\vec V_c = \frac{4}{11} \vec v_0 ## due to momentum conservation during the collision. In that coordinate system: ##\left\{\begin{array}{} \vec v_{He,c} = \vec v_0 - \vec V_c = \frac{7}{11} \vec v_0 \\ \vec...

You can start the problem this way : ##\left\{ \begin{array}{} a(t) = -g_0 \\ v(t) = v(0) - g_0t \\ y(t) = y(0) + v(0)t - g_0 \frac{t^2}{2} \end{array} \right. ## 1) First part tells you: ##y_1(T_1) = 0,\ y_1(0) = H,\ v_1(0) = 0 ## 2) Second part tells you: 2.1) ##y_2(T_2) = 0,\...

Homework Statement A thin target of lithium is bombarded by helium nuclei of energy E0. The lithium nuclei are initially at rest in the target but are essentially unbound. When a Helium nucleus enters a lithium nucleus, a nuclear reaction can occur in which the compound nucleus splits apart...

it makes sense, because of the lack of context, the answer is open to discussion. Now assume that the collision is inelastic and that Q joules are dissipated during the collision. You said earlier that you do not use the diagram to pick a root or another (the case of elastic collision proved you...

wait... For an elastic collision, ##\tan(\theta) = 0## or ##\tan(\theta) = 2##. First option is impossible because momentum is not conserved. Second option gives ##\sin(\theta) = 2\cos(\theta) ##. That gives : ##K_f = \frac{K_i}{9}(\frac{1}{\cos^2(\theta)} + 4) ## By conservation of energy...

Hello, Energy may or may not be conserved, the text does not say, that is why the result depends on Q. The fraction of energy lost (##\alpha##) is a convenient way to link Q with the deflection angle ##\theta##. My result is consistent with perfectly elastic collision: take Q = 0 and keep the...

Sorry, I made a mistake, ##\tan(45) = 1## and not ##\frac{\sqrt{2}}{2}##, but I reach to the same conclusion. Do you agree ?

The diagram suggests that ##\theta## is smaller than 45 degrees. So we should have ##0<t<\frac{\sqrt{2}}{2}##. For exemple, the '+' root is not acceptable for ##\alpha = 1/6##, so I think we have to keep the '-' root. Right?

Homework Statement Particle A of mass m has initial velocity v0. After colliding with particle B of mass 2m initially at rest, the particles follow the paths shown in the sketch (see attachment). Find ##\theta## Homework Equations collisions The Attempt at a Solution The momentum before and...

I see, it would have been simpler to consider the general case ! Thanks !

Yes, there is a recursive way to do it: With momentum conservation and inelasticity: ## mnv_n = m(n+1) v_{n+1} ## So ## v_n =\frac{v_1}{n} ## and total energy lost for n-car collision is ## Q = K_i - K_n = (1 - \frac{1}{n}) K_i ##

Thank you ! What do you mean by 'anything else ?' ?

Homework Statement Cars B and C are at rest with their brakes off. Car A plows into B at high speed, pushing B into C. If the collisions are completely inelastic, what fraction of the initial energy is dissipated when car C is struck? The cars are identical initially. Homework Equations...

oh yes you're right, I forgot the diameter, thank you !

Homework Statement A small ball of mass m is placed on top of a “superball” of mass M, and the two balls are dropped to the floor from height h. How high does the small ball rise after the collision? Assume that collisions with the superball are elastic, and that m<<M. To help visualize the...

I think maximum power occurs as the man leaves the ground because power is a strictly increasing function relative to time: ## \frac{dP}{dt} = (N-mg) \frac{dv}{dt} = \frac{(N-mg)^2}{m} > 0 ## So ## P_{max} = (N-mg) v_{jump} = (N-mg) \sqrt{2g(h_2 - h_1)} ##

yes I was in doubt about that too, I also tried h2 = 4.5 ft which gives an average power of 4.02 hp.

Okay, last part is obtained by replacing ##{v}_{jump}^2 ## by its value, which is given by conservation of mechanical energy between the time the man leaves the ground, and the time he is at the top of his leap. Here, only the weight is working, so ##{v}_{jump}^2 = 2g(h_2-h_1) ##. Let me think...

Hello, thanks for the reply. I don't really understand the question in the problem: what does it mean to 'develop a power' ? My first thought was that I had to find the average power during the time the man had its feet on the ground. Because the forces are conservative, it can be written as a...

Homework Statement A 160-lb man leaps into the air from a crouching position. His center of gravity rises 1.5ft before he leaves the ground, and it then rises 3ft to the top of his leap. What power does he develop assuming that he pushes the ground with constant force? Ans. clue: More than...

I see, thank you very much !

Homework Statement A bead of mass m slides without friction on a smooth rod along the x axis. The rod is equidistant between two spheres of mass M. The spheres are located at x = 0, y = ± a, and attract the bead gravitationally. (a) Find the potential energy of the bead. (b) The bead is...

I agree it is not very realistic :-) Thank you for the reply !

Homework Statement During World War II the Russians, lacking sufficient parachutes for airborne operations, occasionally dropped soldiers inside bales of hay onto snow. The human body can survive an average pressure on impact of 30 lb/in2 . Suppose that the lead plane drops a dummy bale equal...

I get that too, thank you!