Question involving momentum in two carts before and after a collision...?
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.
860183
This is a momentum question.\n\nMomentum = mass x speed\n\nSo, the original momentum is contained solely withing the moving engine and equals 4 units of mass times 5 km/h.  This equals a 'momentum' of 20.\n\nThe final momentum equals the combined mass of the engine and freight car, which is 4 units plus 1 unit for a total of 5 units of mass.\n\nSo, if Momentum = mass x speed, then since our final momentum is 20 (the same as the original momentum) and our final mass is 5, then 20 = 5 x Speed\n\nDividing both sides by 5 yields a final speed of 4 km/h.\n\nBest wishes and good luck.
926868
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
250759
Remember, momentum is affected by mass.  The fly's mass makes it easier to dissipate the inertia.  So it seems that the fly could crash into the front, but not likely. Of course, most people are worrying about other things at the time, so no specific measurements have been taken!  The air inside has mass also, so in effect the car and the air mass have become one unit, in which the fly is also a small part.  When it stops, it helps everything else stop.
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).
969523
if they are both going at the same speed, then the mass,size of the wheel and other considerations have already been take in account, so they both should arrive at the same time.\nGod bless,\ngabe
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.
1074577
We have the same amount of energy in each case, however\n\n\n(A)Kinetic Energy (KE) car1 >0 , KE car2 =0\n50% damage. Energy is being shared equally (since the cars are identical) \n\n(B) KE car1 >0 , KE Wall =0\n100% damage. All energy is observed by the car since the wall is inelastic and immovable.  \n(C) KE car1 = KE car2 >0\n100% damage since both cars had the same amount of kinetic energy.\n\nIn reality the structure of the vehicles and the geometry of collision would determine the extent of damage.
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
87298
It depends on their velocities.  If the velocities are equal then the cart with more mass (the bag filled one) has a greater momentum and therefore would be "harder" to stop.  It would take more energy to bring it to a stop.\n\nObjects in motion tend to stay in motion (and the part everyone forgets) unless they are acted upon by another force.\n\nDo a real world test.  Stand 10 meters downhill from someone holding an empty shopping cart.  When it is released it will roll to you and you should be able to stop it relatively easily.  Next fill one with extra weight and have it roll down to you, you'll probably have to brace yourself or lean into stopping the cart, otherwise it may run you over.  It has more energy and momentum than the other cart.
824337
Assuming that the boat is not moving under constant power, and neglecting the force of water resistence, the boat's new velocity would be 9.33 m/s.  \n\nWe know this because of conservation of momentum.  The momentum after the collision must be the same as the momentum before the collision.  Since momentum is equal to mass * velocity, the boat's momentum before the collision is:\n\n510kg * 11m/s = 5610kg*m/s. \n\nSince the momentum is the same after the collision, we just divide the total momentum by the new combined weight to get the new velocity:\n\nV = 5610kg*m/s / (510kg + 91kg) = 9.33m/s  \n\nThe speed with which Batman hit the boat does not factor into the equation, unless we're interested in the boat's angular velocity.  I'm assuming that you only want to know the boat's horizontal velocity over the water.  Since Batman had no horizontal momentum (he was falling straight down), his movement would not affect the horizontal motion of the boat.
771612
hello\n\na)take the momentum of the sports car and divide by the total mass of  both cars. 25×920÷3220=7.14m/s\n\nb)using the same method as a), we can find the answer. but use a little logic here, if the total mass doubles then the final speed must be halved for conservation of momentum to be followed. v=12.5m/s\n\nc)KE=0.5mv². speed is halved. v² means kinetic energy should be quartered (0.5²). but mass was doubled due to there being 2 cars now. so overall kinetic energy has only been halved (0.5²x2). the other half is lost.
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.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
803571
For the first question: momentum is equal to mass times velocity, therefore; if the motorcycle has a higher velocity then its momentum would be equal. Keep in mind the momentum could only be the same if the velocity of the motorcycle is greater than the velocity of the car by the same factor that the mass of the car is greater than the mass of the motorcycle. \n\nex.  car: \nm=1200 kg\nv= 20 m/s\n\nmotorcycle: \nm= 400 kg                             \nv= 60 m/s\n\np= momentum\np(car)= 24000 kg x m/s\np(motorcycle)= 24000 kg x m/s\n\nFor your second question I would have to guess that if you spin the sack in a centripetal motion and stopped it at a certain point, the tangential velocity would propel you in the desired direction. Think about a fast pitch softball throw using the sack without letting go. In this case the radial acceleration would be conserved as linear acceleration which would propel you in whichever way you want to go.
896561
If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
823386
Hi Alan the 1st!\nWe neglect the central aperture in CD, then inertia momentum is I=m*r^2/2 for any homogeneous disk, m=0.017kg, r=0.06m. Torq T=I*w’, w’ is angular acceleration, while w=w’*t, w=22 rad/s, t=0.88s, thus T=m*r^2*(w/t)/2, - plug in the proper numbers, be a good boy!