How fast must a particle move if its momentum is to be 3.3 times its classical momentum?
551506
I'm assuming you are talking about a relativistic particle here.  Correct me if I'm wrong.\n\nClassical momentum is p=mv\n\nMomentum in relativistic mechanics is p=(gamma)*mv where \n\ngamma= 1/sqrt(1-(u/c)^2)  \nu=speed of particle\nc=speed of light\n\nNOW, what you are saying is:\n\n(gamma)*mv=3.3*mv\n\nNotice m and v cancel and you are left with:\n\ngamma=3.3\n\n1/sqrt(1-(u/c)^2)=3.3  \n\nInvert both sides then square:\n\n(1-(u/c)^2)=0.091827\n\nSolve for u (skipping some steps b/c you should be able to do algebra by now):\n\nu=0.95298*c\n\nThe answer:  95.3% the speed of light\n\nOh yeah!
566668
No object that has mass can be accelerated to move at the speed of light. If you were somehow moving at 0.99c and then shown a flashlight, the velocity of the photons would not be 1.99c, it would just be c. The speed of light is invariant. The true formula at relativistic speeds is the following:\n\nV = (v+u) / (1 + vu/c²)\n\nWhen the velocites are not relativistic (i.e. throwing a ball out of a moving car) this approximates to V = v+u. So if you are moving toward something that is emitting photons at c, and you yourself are somehow also moving at c, the velocity you observe the photons to be moving is still just c. Try it in the formula: V = (c+c) / (1 + cc/c²) = c\n\nIt can be derived from special relativity or the Lorentz transformations.
371210
3 * 10^8 m/s (approx.- this is used for the purpose of calculations)\nthe exact value is hard to calculate but the closest approximation is 299,792,458 m/s
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
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.
829059
I hope I did this correctly.  Would you let me know?\n\nI set up the equation of the relative distance the particles will have from eachother at time t:\n\ndistance equals location of particle one minus location of particle two:\n\nx=\n(4t-3)i+\n(6-3t)j+\n(5t)k\n\nthen I distributed the i, j and k to get\n4it-3j+\n6j-3tj+\n5tk\n\nThen I took the derivative dx/dt treating i, j, and k as constants to get\n4i-3j+5k\nthis is the minimum separation\n4i-3j+5k\n\nThis I then used to set up the following three simultaneous equations to find the time that occurs:\n\n4t-3=4\nt=1\n\n6-3t=-3\nt=1\n\n5t=5\nt=1\n\nj
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.
83134
you really should have given numbers \n\nkinetic energy will be 1/2mv^2\n\nIts linear speed can be determined by distance over time  (d/t)\n\nthe next 3 answers need me to know what data you have in order calculate their values
852502
Momentum is a measure of mass times velocity, so just multiply them together:\n\n250kg * 6m/s = 1500 kgm/s\n450kg * 2m/s = 900 kgm/s\n\n1500 > 900, ergo the Dolphin has more momentum
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
357903
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
1120774
WHAT IS M\np (momentum)=m (mass)* v(velocity)\nsquare root of momentmum will be the product of the square root of mass into square root of velocity
878067
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
787981
let al' mass be m\n\nconserving momentum\nm*0.91=(158-m).829\nsolving for m we get\n         m=75.320kg
544447
c represents the speed of light in a vacuum.\n\nc = exactly 299,792,458 m/s