An elementary particle moving at 1.35 x 10^5 m/s strikes an identical particle initially at rest. After the?
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.
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.
1155452
Only impulse (force x duration, or the quantity of momentum transferred to the pin) is calculable from the data, not force by itself, which is variable in time and not directly meaningful.\n\nIf the moving object has mass M and velocity V, and the object being struck has mass m, then assuming the collision is elastic and all motion is in the forward direction (all these assumptions are at least approximately true for the bowling ball and pin) then the impulse is:\n\nImpulse = 2*M*m*V/(M+m)\n\nEach pound value must be divided by g to get proper units of mass, so this is:\n\n=(g/g^2)*[2*16*3.5/19.5]*V = 5.74*(V/g)\n\nSince:\n\nV/g = 23mph/(32ft/sec^2) = 1.05 seconds \n\nthe result is:\n\n6 pound-seconds\n\nIt's the equivalent of 6 pounds acting for 1 second, or 1 pound acting for 6 seconds, or etc..
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.
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.
212614
the wall stops it and the ground starts it again. if the spinning force is greater than the stopping force it would continue to spin after hitting the wall.
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
749375
You mean at the same rate - 32.2 feet per second per second.\n\nGravity pulls on all matter at the same rate. If you drop a bowling ball taped to a baseball, they will fall at the same rate. But it is not because the bowling ball is pulling on the baseball, it is because gravity is pulling on both of them. If the tape breaks gravity will still be pulling on both of them at the same rate. They will hit the ground at the same speed, however, they will hit with different forces. Because Force = mass x acceleration. The acceleration is the same for both of them. But the mass of the bowling ball is larger than the mass of the baseball, so it will hit the ground with more force.
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
1090203
Its not just for a microscopic particle. Its for any particle. The effect is only large for very small particles though.\n\nThink of it this way.\n\nIf you want to know precisely where a particle is you will have to use a very short wavelength of light to measure its position - you cannot be more accurate than the wavelength of the light. But very short wavelength light carries a lot of momentum. Each photon has momentum h/wavelength (h is Planks constant) - the shorter the wavelength the greater the momentum.\n\nBut if you hit the particle with a photon with a lot of momentum, it will accelerate it. As a result you will not be able to get an accurate measurement for its momentum. To measure its momentum accurately you would have to use a very long wavelength photon.\n\nThese are competing requirements.\n\nThe Hesienberg relationship quantifies these competing requirements by saying that the error in position multiplied by the error in momentum cannot be lower than hbar (Planks constant over 2 pi).\n\nPlanks constant is very tiny - around 6.62 e-34 Js. So it only matter for tiny particles.
83355
Okay, you know the right hand rule. Good.\nFirst consider the effect of only the e-field: The e-field is pointing in the +x direction. Therefore, the force it exerts on the proton will also be in the +x direction.\nSecond, consider the effect of only the magnetic field: The magnetic field is pointing in the +y direction; therefore, the force it exerts on the proton will be in the -x direction (you use the rule you described to figure this part out). It so happens, that in the first scenario, the force exerted on the proton by the magnetic field exactly balances that exerted by the electric field. Therefore, the movement of the proton is unaffected.\n\nHowever, in the second case, the proton is accelerated through a potential difference of 2V. Therefore, it will be moving faster (think about it: a higher potential difference has greater "potential" to pull the proton.  Also, the equation F = qvB where q is the charge of the proton, v is its velocity, and B is the magnetic field strength, states that the faster the proton is moving, the greater will be the force exerted on it.  So, compared to the first case with only a 1V potential difference, the magnetic field here will exert a greater force on the proton. But, the force exerted by the electric field is unchanged. So, the net force on the proton will be in the -x direction.
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.
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
544447
c represents the speed of light in a vacuum.\n\nc = exactly 299,792,458 m/s
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
124894
a push wave in which the particles move in the same direction to that of the wave as opposed to transverse wave in which they move vertical to the wave direction