Two Similar objects with different weight falling at the same speed . (Revised)?
612059
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
201969
Answering...\n\nExperiments have been done in places where air resistance is negligible...and it was found out that objects with different masses have the same acceleration as long as the test are done within a reasonable height...Therefore, the answer is C.\n\nThis can be shown in equation form, formulated by I. Newton:\n\nF = m x a ... where:\n\nF = force in newton\nm = mass in kg.\na = acceleration in m/sec^2\n\nSo:\n\na = F/m = 2 N/2 kg = 1 m/sec. squared
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.
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
1103152
Falling objects do not fall at a constant speed, they accelerate (get faster) the longer they fall. In a vacuum (no air resistance) two objects of the same mass (weight) experience the same acceleration. So if they're dropped from the same height, they will hit the ground at the same time. \n\nIn the atmosphere, air resistance will diminish the force of gravity, so two objects of different shape or density would accelerate at different rates because of different air resistance.
195717
dba_gregg has missed the point that both are same weight\n\nThe moment of Inertia of a hollow sphere  is 1.6 times greater than the solid sphere and so it rolls down slowly.\n\nThe speed of solid sphere will be 1.09 times the speed of hollow sphere.
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
425701
No. You can understand the difference by going back to its source: the Newton's Second Law.\n\nF = ma\n\nThe weight of an object is the attractive force (F) exercised by the Earth on that object.\n\nAs the above equation shows, the mass (m) of an object is the constant of proportionality that links any force exercised on the object with the consequent acceleration (a).\n\nSo you see that the weight is a special case of force, that you have when a is the Earth's gravity acceleration. \n\nNow suppose you bring the same object on another planet, say Mars, where the gravity acceleration is different. Its mass doesn't change, but the weight does!\n\nThe weight of an object is an Earth accident, while its mass is a more fundamental concept because its value remains constant (at least within the non-relativistic limit, but this is another story).
874570
NO!!  If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.\n\n1.  If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other.  So ALL of the moving ball's K.E. has been transferred to the other ball.  \n\nHere are two ways to see this:\n\n1a.  The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this.  Most undergraduate physics instructors must haul this out from time to time.\n\n1b.  You can easily solve it analytically, but a thought experiment will also demonstrate it.  In the originally stationary ball's frame, let the other ball approach with speed v.  Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)!  Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2.  Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v.  (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)\n\n2a.  A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary).  Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and  MV.  They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds.  Then you can do the other frame-jugglings from there. \n\nHowever, this analytical approach, done accurately, is a tedious mess.  So, to appreciate the general problem more physically, consider just the following:\n\n2b.  The case of m << M.\n\nConsider the "equal but opposite momentum" frame for these two balls, in which MV = mv.  This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.\n\nSo, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v.  (In fact identically so.)  This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2).  (There's no need to do picky constant book-keeping for these purposes.)  \n\nNow after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.)  But, being loose about the constants, M's K.E is only O(MV^2).  \n\nSo, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)\n\n= (m/M)(M/m)^2 (since mv = MV)\n\n= M/m.\n\nBINGO!  The little bugger still carries by far the larger amount of the kinetic energy.  The massive ball (i) moves sluggishly, and (ii) because  K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.\n\nThis is a well-known fact in particle collision theory, that little particles basically bounce off big ones.  \n\nYou could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.  \n\nIt's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
1149825
I'll give you a head start, and we'll see if you can figure it out from there:\n\nFirst, the givens:\n\nd = height from the ground to the point at which the two balls are released = 122.5 meters\n\nv1 = initial velocity of the first ball = 0\nv2 = initial velocity of the second ball\nt1 = time ball 1 is dropping (seconds)\nt2 = time ball 2 is dropping (seconds)\nt2 = (t1)-2 seconds\n\na = gravitational acceleration = 9.8 m/sec^2\n\nFor both balls:\n\nd = vt + at^2\n\nFor the first ball, v1 = 0, so d = at1^2\n\nFor the second ball, d = v2t2 + at2^2\n\nThe distance in both case is the same, so:\n\nat1^2 = d = v2t2 + at2^2\n\nOkay, so here you go:  You have the basic equations,so solving is trivial arithmetic:\n\nd = 122.5 meters = (9.8 m/sec^2) (t1)^2\n\nand solve for t1.  Once you know t1, then you know that t2 is two seconds less, so with that value you can solve:\n\nd = 122.5 meters = v2t2 + (9.8 m/sec^2)(t2)^2\n\nI leave the rest to you!  (And I have calculated the answer if you want to check it)
578701
Both are Same Weight , 1 Kg\nbut Volume is different
449872
The both weigh the same
794045
we have F= G*M*m/r^2  and F=weight\n\nwhen M=M and r= r/2 then F=G*M*m/ (r/2)^2\n\ntherefore F(new)= 4*F and hence new weight is 4 times.\n\nthis is due to change in the value of g(acceleration due to gravity)\n\nsince F=mg= G*M*m/r^2   and hence only g is effected\n\nand hence weight also changes mass remainig same.
524636
If the forces are both acting in the same direction, add them together. = 65 Newtons
744356
I'd say (A) because momentum is directly proportional to mass.  The smaller the mass, the smaller the momentum.\n\nMomentum is the product of mass and velocity\n\nsince the force is equally applied to both masses, their velocities will be the same\n\nthe only difference are the inertial forces generated, which would be less for the smaller mass and more for the larger mass