What are the parameters that change during collision between masses?
816105
Generally, some of their collective kinetic energy is converted into heat, so the KE after the collision is less than the initial KE.  Momentum is conserved.
714752
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second
966112
u can consult ur physics book
865948
KE = 1/2 mv^2\nfind the KE before the collision using the formula.\nfind the KE after the collision using the formula.\nsubtract the two values to find the change in kinetic energy
769961
Hi, \n\nIn general, the equation describing the position of an object under the influence of a constant force is as follows:\n\nh(t) = at^2 + vot + xo \n\nwhere : \nh(t) = height at any time t\na = acceleration\nvo = initial velocity\nxo = initial height.\n\nSo the answer would be \nF= m*a (mass times accelleration) = (the mass kg)*(1.96m/s^2) \n\nI'm guessing that you want to find out the equation for acceleration, so what you should do is use the values for t, and   h(t) to solve for the constants (xo, vo)\n\nexample: \nat time t=0, height is 0.50m\n\nso h(0) = a(0)^2 + vo (0) + xo = xo\n\nbut that also equals 0.50m;\n\nso:\nxo = 0.50m.\n\nand the equation becomes \nh(t) = at^2 + vot + 0.50m. \n\nThen use the different values of t, and h(t) to do the same thing with the other constant, and you'll have it. \n\nHope that helps, \nMatt
296792
Your experimental model is corrupt in terms of current known information about matter and energy in a space-time continuum.  No prediction or calculation can be applied.\nWith two point masses, the only force in the system would be do to gravity.  This is an assumption based upon multiple moving masses in what we would call the normal physical universe.\nIf the distance between their centers (d) was sufficiently small, the points would be touching one another and no force would be exerted on the masses.  If the distance was sufficiently large, then it is possible that the force would be so small that the masses would not move.  Only in a finite range of distances would any gravity possibly exist.\nOne can make no assumptions about G as a gravitational constant, since G assumes that one of the masses is the earth.  One would need additional information about the masses to solve the problem.
946504
Intensity varies as the square of the distance.\nThat should help
322347
it is the collision of two preasure sysems and the difference of the tempeture
109475
When masses collide, they exchange energy and momentum.  I'm not sure what you mean specifically by "power division", but you might be referring to this exchange of energy, since power is defined as energy transferred per unit time.  If so, just know that the important thing is that the total amount of energy and momentum is conserved, that is, remains constant.\n\nAddendum-  What the next correspondent says about elastic and inelastic collisions is correct; it makes a difference in how energy is transferred.  In every collision, total energy is conserved, as I said, but in elastic collisions, mechanical energy is conserved as well.  That is, none of the energy is converted to other forms of energy like heat, light or potential energy of deformation.  In inelastic collisions, momentum and total energy are conserved, but some of the mechanical energy is converted to those other forms.
709931
kaksi_guy: I think charlesellis753 is correct.  If we interpret the question as a fill in the blank, mass appears to be the best answer.\n\nIf the ______ of an object doubles, then the force required to accelerate it at the same rate will double.\n\nThat sentence is true is the blank is filled with 'mass'.
860846
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
791190
Only if E=MC^2\n\n... it follows that if there is a change in the rest-energy of a body, there must be a corresponding change in its rest-mass. For example, if a body is heated, and thereby absorbs a small amount of energy ΔE (as measured in the frame of reference in which the body is at rest), its rest-mass will increase by a very small amount equal to ΔE/c2. This increase is tiny because of the high numerical value of the speed of light. Indeed, for mid-sized objects, such an increase in mass would be too small to measure with even the most accurate balance. For example, if a 1 kg block of gold is heated so that its temperature increases by 10 °C, then, ceteris paribus, its mass would increase by as much as 1.4 × 10-14 kg; a cube of gold of this additional tiny mass would have sides smaller than one one-thousandths of a millimeter. Similarly, if a body emits an amount of energy ΔE, say in the form of light or heat, its rest-mass will decrease by a tiny amount ΔE/c2. In both cases, the important and novel claim made by SR is that the inertial mass of a body can change depending on whether it absorbs or emits energy. This is true, of course, not just for isolated mid-sized objects but for any physical system.
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
759179
This can be describe using kinetic and potential energy.\n\nMvi1 + 3Mvi2 = Mvf1 + 3Mvf2 (M colliding with 3M, giving v final being v for 3M)\n\nvf=1.66m/s (final velocity of 3M object after collision)\n\nU=K (Potential Energy equals kinetic when it reaches max height))\n\nh(3M)g=1/2(3M)vf^2\n\nh=0.283m\n\n.5 - h = 0.217m\n\narccos(0.217/0.50) = 64.3 degrees from vertical
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.