How can the total momentum remain constant if two objects collide, retain their mass and stop moving?
546536
Take two objects of the same mass moving towards each other (in opposite directions) with the same speed. Let's say one object has momentum of p and the other -p. The total momentum of the system (p-p) is 0. When they collide, they both stop moving and the total momentum of the system is still 0.
716426
If the objects are perfectly elastic (that is they don't deform), then you can use the laws of conservation of momentum and conservation of energy to solve the problem. If they (or one of them is) are plastic...that is they deform during the collision, then conservation of momentum still applies, but it is more difficult to use conservation of energy unless you know how energy is lost in the deformation process.\n\nI don't mean that to sound complicated, but that's it in a nutshell.
742829
Momentum is conserved. Figure out the total momentum of all three carrianges before the collision, taking into account the direction of the momentums.\n\nAfter the collision, the momentum of the three linked carrianges must be equal to the momentum of the system before the collision.\n\nIf the other two carriages are not moving before the collision, then the velocity of the linked system will be 40/3.
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
581268
The mass of any body changes with velocity.\n\nIf the masss of the body is m0 when it was rest, then the mass ,when the velocity is v, is  m =m0/ square root of {1- (v/c)^2}.\n\nThe momentum is  equal to mv which is greater than mov
839596
Consider the impact as elastic impact.\n\nHence the kinetic energy is conserved.\n\nTherefore, the magnitude of the velocity cannot change.\n\nThe angle of incidence of the sphere is equal to the angle of reflection.\n\nThe initial momentum normal to the wall is mv sin a.\n\n= 3 x 10 x sin 60 = 25.98 kg.m/s.\n\nAfter impact the momentum is - 25.98 kg.m/s.\n\nChange in momentum is 2 mv sin a = 2 x 25.98 kg.m/s = 51.96 kg.m/s.\n\nForce is rate of change in momentum = change in momentum / time of impact.\n\n= 51.96 / 0.2 newton.\n\n = 259. 8 newton.
42744
momentum is speed and mass is the substance of objects
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.
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
95497
This a momentum question.\n\nRemember momentum is:\n\nm * V\n\nm = mass\nV = velocity\n\nSo the conservation of momentum is:\n\n(m1 * V1) + (m2 * V2) = (m1 * V3) + (m1 * V4)\n\nMomentum (before collision) = Momentum (after collision)\n\n(Notice mass does not change after the collision, but velocity does.)\n\nSo let's plug in the variables:\n\nm1 = 2.0 kg\nV1 = 3.5 m/s\n\nm2 = 4.5 kg\nV2 = -5.0 m/s \n(negative indicates object is traveling to the left)\n\nV3 = ?\nV4 = -1.8 m/s\n\n(2.0 * 3.5) - (4.5 * 5.0) = (2.0 * V3) - (4.5 * 1.8)\n\n(7) - (22.5) = 2(V3) - (8.1)\n\n-15.5 = 2(V3) - (8.1)\n\n-7.4 = 2(V3)\n\nV3 = -3.7 m/s \n\n(the negative sign just indicates direction with relation to the axis; in this problem, negative indicates the object is traveling to the left)\n\nSo Object A will be traveling to the left at 3.7 m/s after the collision.
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.
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).
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
115093
Momentum is the tendency of an object in motion to stay in motion. The speed of light is an actual speed.
357903
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
916585
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s