if a mass collides  at  large mass a force oc curs because the mass was blocked.would it still have momenturm?
486047
If after collision the mass goes to rest i.e. no velocity then momentum is zero.
156917
It depends on how it came to rest. If it had Striked any other ball then its K.E will be transfered to the other. If u were to catch it , then imagine how u will bring your hand towards the ground so as to reduce its impact.Thus its k.E is transferred to you.\n               But if it were to strike the ground then the case is different.From conservation of momentum ,M1V1=M2V2.If M2 be the mass of earth, it is very large when compared to that of the ball. So to satisfy the above condition, V2 should be negligbly small. so we do not face amy impact. Now consider the case of an asteroid or an Hydrogen bomb.We r sure of expriencing the impact.
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
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.
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'.
525497
Of course yes. Inertia is described by the first Newton's law of motion, which applies to any mass, be it a rigid body or not. You can imagine this experiment: put on the bus floor a glass sphere (a good approximation of a perfectly rigid body) and imagine what happens when the bus deviates from resting or moving uniformly: the sphere moves w.r.t. the floor. Its motion is not related to its rigidity, but to the fact that the sphere is not connected to the bus structure, therefore it doesn't receive any external force contrasting its inertia.
854814
My magic 8 ball says no. \n\nSo does my physics professor.
994962
Mass is kind of like the object's inherent inertia.  The inertia it would have regardless of its location and even if it were standing still.\n\nAn object's inertia is it's resistance to change.  Different things besides mass can affect an object's inertia.  For example, it's position can affect inertia, especially when talking about rotational motion.  An object located at a 1 meter wheel's rim has more inertia than an object located at the wheel's hub (lots more, in fact, since moment of inertia is equal to mass times the square of it's distance from the rotational axis).
407008
Only non-fluids will keep their shape.  If the energy of the constituent particles is not sufficent to overcome the bonding then the mass will keep its shape unless a large force is applied to it.  If the gravitational force is large enough it will most certainly brake objects; would you like to try walking past a black hole?
279767
As far as we can tell, space is infinitely big.  It's rare for galaxies to collide.  It does happen, however.  Below is a link to an article about a recently seen collision.  Keep in mind, however, that because of the distances involved these galaxies actually collided millions of years ago, and the light is just reaching us now.
68476
Energy equals mass times the speed of light squared. It means that mass contains a huge amount of energy, which is why just a tiny amount of plutonium results in a huge nuclear explosion.
300802
if the point is big the gravity don't effect it much but if it is small gravity will effect it i forgot the whole explanation though \n\nyou could try this link
771568
Basically you have the right idea. Use momentum (mass x velocity) for the motion and the sudden change in speed and/or direction (acceleration) and you get force x time = kinetic energy trying to rip you apart or squash you like a bug.
193418
Heisenberg notwithstanding, I have never observed my chair turn into energy, even as a transient state, no, not even for an instant.  It must be when I'm not looking.\n\nEnergy, if unused as work, turns into mass... in my case, a mass of fat around the middle.
299265
The answer has to be A.\nThe time will be the same because Gravitational force does not depend on the mass of the object, since they start with same velocities and accelerate with the same rate they must hit the ground at the same time.\n\nAnd it obviously has to be the same distance because if it wasn't then they would never hit the ground at the same time in the first place. That would happen because if 2 bodies start with the same speed and have the same acceleration but different displacements then their time has to be different.
146188
depends on the activity and on the mass
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