1.Suppose you place a ball in the middle of a wagon and then accelerate the wagon forward.?
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You physics types ALWAYS have to delete something, eh? Like mass, or air resistance or friction! lol\n\nAnyway, I think the ball would be a) either moving forward or stationary, depending on the wagon's rate of acceleration, relative to the ground, and b) rolling backward (due to inertia), relative to the wagon.
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it depends how fast u go. lol
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use conservation of energy:\n\nThe 810 N force did work for 2 m\nthe result is potential energy gained in height working against gravity including the 2m that the 810N force acted on the ball\n\n810*2=7*9.8*(h+2)\nh=1620/(7*9.8)-2\n=21.6m\n\nj
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This is related to Newton's law: An object in motion or at rest will stay in motion or at rest unless acted upon by an external force.\n\nTo elaborate, an object travelling in a straight line will keep on travelling in that straight line at a constant speed unless something acts on it.\n\nWhen you are in the car with your ball, everything in the car is travelling at the same speed, so you through the ball up and it comes straight back down. But if you throw the ball up and then jam on the cars brakes, the ball will keep moving at its original speed but the car is slowing down, so the ball would fly forward into the front windshield.\n\nLast point, if you throw the ball out the window, it will appear to fly backwards - It is just slowing down compared to the car because unlike the air inside the car, the air outside is at rest, so the ball will be flying through the air at a high speed and the wind drag will slow it quite rapidly.
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possibly, number of "wheels on a wagon"
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Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
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a=F/m, lim x->inf x/x = 1. The object will accelerate.
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a) they give you everything you need.  v0 = 32, s0 = 0, so\n\ns(t) = -16t² + 32t\n\nb)\n\ns(1) = -16 + 32 = 16\n\nc)  The ball hits the ground when s(t) = 0 (the second time).\n\n0 = -16t² + 32t = -16t(t-2) so t=0 or t-2 = 0, t=2.\n\nSo the ball hits the ground when t=2.\n\nd)  The maximum height is attained at the vertex of the parabola.\n\ny - 16 = -16(t² - 2t + 1)\ny - 16 = -16(t - 1)²\n\nSo the vertex is (1, 16), that is, at time t = 1, when the height of the ball is 16.
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1.  Hurl it spinning into space.  The vehicle will spin around its center of gravity, and its center of gravity will travel in a straight line.\n\n2.  Balance it on a see-saw three ways: upright, on its side, and on one end.  Each balance will define a middle plane that cuts through the vehicle.  The intersection of the three planes is the center of gravity.
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Let's assume that the cue ball is traveling in the +x-direction\n\nConservation of energy says\n\n0.5*m*v1^2 = 0.5*m*v2^2 + 0.5*m*v3^2\n\nwith the condition that all the masses are equal this reduces to\n\n(1): v1^1 = v2^2 + v3^2\n\nI'll assume that v2 is the speed of the 8-ball and v3 is the speed, after collision, of the cue ball.\n\nConservation of momentum says\n\nm*v1 = m*v2*cos(q) + m*v3*cos(a)\n0 = m*v2*sin(q) - m*v3*sin(a)\n\nfor the x and y directions, respectively. Here q is the angle that the 8-ball moves away at (35 deg) and a is the angle that the cue ball moves away at. These reduce to\n\n(2): v1 = v2*cos(q) + v3*cos(a)\n(3): v2*sin(q) = v3*sin(a)\n\nWe now have the three expression (1), (2), and (3) for which to solve we can solve for the angle a.\n\nFrom expression (3)\n\nv3 = v2*sin(q)/sin(a)\n\ninserting this into equation (2)\n\nv1 = v2*cos(q) + v2*sin(q)/tan(a)\n\nTaking these two expression and plugging them into equation 1.\n\nv2^2*(cos(q) + sin(q)/tan(a))^2 = v2^2 + v2^2*(sin(q)/sin(a))^2\n\nfrom which we can eliminate v2 to get and expression just in terms of the angles. Expanding this expression we have\n\ncos(q)^2 +sin(q)^2/tan(a)^2 +2*sin(q)*cos(q)/tan(a) = 1 + sin(q)^2/sin(a)^2\n\nrearranging terms and using sin(x)^2 + cos(x)^2 = 1 we have\n\n-sin(q)^2 + sin(q)^2/tan(a)^2 + 2*sin(q)*cos(q)/tan(a) = sin(q)^2/sin(a)^2\n\nmultiplying by sin(a)^2\n\n-sin(q)^2*sin(a)^2 + sin(q)^2*cos(a)^2 + 2*sin(q)*cos(q)*sin(a)*cos(a) = sin(q)^2\n\ncollecting terms\n\nsin(q)^2*(cos(a)^2 - sin(a)^2 - 1) + 2*sin(q)*cos(q)*sin(a)*cos(a) = 0\n\nwhich becomes\n\n-2*sin(q)^2*sin(a)^2 = -2*sin(q)*cos(q)*sin(a)*cos(a)\n\nseparating the a and q terms and simplifying\n\ntan(a) = 1/tan(q)\n\nwhich is readily solved for the a, the angle at which the cue ball moves after the collision.
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B.  The acceleration is towards the Earth and the force is constant.
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The ball's vertical acceleration will constantly decrease due to the gravitational force. This happens until the ball reaches the apex of it's path, then will start to accelerate constantly back down. This vertical acceleration happens independent of the horizontal path.
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the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s
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If the vehicle is moving at a constant speed, then the object is moving forward at that same speed before being tossed. After it is tossed, it continues to move forward at the same speed (as you and the vehicle) because there is nothing to slow it down.\n\nTry tossing the object when the vehicle is accelerating. The object, after being tossed, does not enjoy the acceleration that you and the vehicle do, but instead continues to move forward at the speed it had when tossed, and will land "behind" you. This is the principle behind a famous result of Einstein. If the vehicle has no windows, you can't tell whether the vehicle is accelerating, or standing still with "gravity" pulling tossed objects toward the back of the vehicle. Even if you try to use a beam of light to determine acceleration or gravity, you will get the same bending of the light beam.
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1. KE = (1/2)*m*v^2 - you have the KE and mass.\n\n2. The amount of work done will be equal to the kinetic energy the arrow is given. \nW = F*d = (1/2)*m*v^2 solve for v.\n\n3. When the speed is increased 50% Vnew = 1.5*Vold. Kinetic energy will increase by 1.5^2 = 2.25. \n\nWork done stopping it is Fbrakes*distance and equals the kinetic energy it has. \nKEnew = 2.25*KEold so the stopping distance, SD, changes accordingly:\nSDnew = 2.25*SDold\n\n4. The acceleration on the baseball is given by\nV^2 = Vo^2 + 2*a*d where V is final velocity, Vo is original velocity, a is acceleration, and d is distance. Finally, \nF = m*a.\n\n5. \n(a) The tension is divided between holding the weight and accelerating it. T = W + m*a.\n(b) The work gives the load a potential energy. \nU = m*g*y\nIt also gives it kinetic energy. The velocity it has at the 21.0 m height is given by\nV^2 = Vo^2 + 2*a*y. Use the velocity in the KE equation. The work done is the sum U + KE.\n(c) Might be a trick, but this feels like semantics. I don't see that \nb & c are any different. Maybe in class, or the book, some distinction was made between those 2 ways to say it?
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You know it is harder to pull a wagon loaded than empty and a lighter wagon is easyer to pull than a heaver one. What happens as the particle gets less mass with the same push. How fast does a particle go with no mass?? Infinity would be silly so they go at speed of light.\nYour spaceship must be less than the speed of light and when you run it is not simple addition as Newtonian Mechanics,..
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1)Start by modelling the speed of a ball in one direction; take into account friction force (negative drag) that is equivalent to negative acceleration. Loop until speed is equal or less than to zero. Then, stop.\n\n2) Do Exactly the same, now in a 2-D space: model how'd the would move given an initial vector speed.\n\n3) Model a "Hit" on a ball: given an instant (vector) speed plus a "Hit Vector", compute the resultant vector; then use that vector as the initial speed again.\n\n4) develop a "Hit?" function that returns true if the difference between the centers of the balls is less than the diameter, or if the center of the ball is closer than a radius to a wall (the stick can be thought of as another ball)\n5) Every time Hit? is true, recompute vector speed for every involved ball.\nVoila! you have your game. Let me know if it's useful and the results.