a 1kg ball is dropped vertically on the floor with 25m/s.it rebounds with initial sspeed of 10m/s.?
446638
impuse is the change in momnetum which is mvfinal - mv initial\n\ntherefore 10*1 - (-25)*1 = 35 kg/ms. Took down as minus therefore 25ms becomes minus in the equation.\n\naverage force is therefore just the impuse divided by the time of contact and is therefore \n35/0.02 = 1750N
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
862752
This question deals with either conservation of enregy, or conservation of momentum. In this case it will be easier to go with momentum, so that's what I'll do.\n\nThe momentum before an event will equal the momentum after the event. In this case, the momentum before the event equaled zero, so the momentum after must equal zero also.\nIf the ball has a momentum of (.350kg)(28m/s)=9.8kgm/s, then the man must have an equal and opposite momentum to cancel it: (65kg)(v)=-9.8kgm/s. Since we now know the mans mass and momentum, figuring out his velocity is simple:\n(65kg)(v)=-9.8kgm/s\nv=-0.15m/s
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
579002
First I'll assume this isn't a trick question, i.e., that the balls start with the same speed and have equal densities. They are rigid and roll on a rigid surface. What slows them down? Air resistance and going uphill. Not friction because they are rolling, not sliding. \nWith the same initial velocity, they have the same translational momentum per unit mass. However, another factor is the angular momentum due to rolling. Let ball 1 have a radius r=1 and a mass m=1, and ball 2 have r=2 and consequently m=8. If each ball has an initial velocity of 1, ball 1's angular rate w=1 rad/sec and ball 2's w=0.5 rad/sec. The moment of inertia I of a solid sphere = k*m*r^2 (k=0.4), or I=k for ball 1 and I=32*k for ball 2. The angular momentum = I*w, thus I=k for ball 1 and I=16*k for ball 2. While both balls have the same translational momentum per unit mass, ball 2 has twice the angular momentum per unit mass and therefore a greater total momentum per unit mass, which means a greater force * time per unit mass is needed to stop it. Going uphill provides the same force per unit mass on each ball, so ball 2 goes further uphill. Air resistance per unit mass is greater for ball 1 since air resistance depends on area, and ball 1's area/mass ratio is 1 vs. ball 2's ratio of 0.5. Again, ball 2 goes farther.\nEDIT: The balls have different momentum per unit mass but the same energy per unit mass, so my argument above about uphill travel is wrong; they go the same distance uphill. But air resistance still favors the larger ball. So answer 5 is the best answer so far. Another real-life factor would be: suppose the balls are on a rough surface or face an effective drag of a surface such as a rug. A larger ball wins on both. The larger radius means smaller vertical motions for a given scale of roughness (e.g., gravel size), so less energy wasted per unit mass. And you can visualize how a deep-pile rug would stop a rolling golf ball much faster than a bowling ball.
964449
Depends on what mass, what size due to air resistance\nA ball bearing coming from orbit could be in excess of 5mps, a man falling (not from orbit! but maybe 50,000') about 180mph.
896193
This is an impulse/momentum problem.  Impulse is ∫f*dt[0 - t1] it must equal the change in momentum.  Impulse is given by Favg*t, in your case 0.012*16000 N*sec  Therefore the momentum of the student on impact was 0.012*16000 = m*v.  m is the mass of the student, 63kg.  Therefore his velocity on impact was v =0.012*16000/63.  He fell under acceleration g = 9.8m/sec^2.  Velocity after falling time t is g*t.  Distance traveled under uniform acceleration is h=.5*g*t^.    Use the velocity to get t, then compute h.
409505
A soccer ball of mass 0.43 kg leaves the foot of the kicker with an initial speed of 25 m/s.The weight of the ball at the start of the game shall not be more than 450 g nor less than 410 g.\nSource-Sam
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..
904552
Both balls would hit the rocks at the same speed.\nBall #1 would leave your hand travelling straight up at speed 'u'. It travels up, stops, and starts plummeting back to earth. When it reaches the height at which it left your hand, it is travelling at speed 'u' straight down. The speed that it hits the rocks at is u+(time x 10m/s/s).\n\nBall #2 would leave your hand travelling straight down at speed 'u'. The speed that it hits the rocks at is u+(time x 10m/s/s). It is the same speed.
815859
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
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.
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
786359
In theory, it will never stop, so I'll assume that when it bounces less than 1 mm it is finished.  Total distance travelled will be 26.99221155 meters.  Start with the 3 m drop and add the 2.4 m bounce and drop and continue the sequence until the bounce is less than 1 mm.  Here's my spread sheet:\n\n3000----------3000\n2400----------7800\n1920---------11640\n153614712\n1228.817169.6\n983.0419135.68\n786.43220708.544\n629.145621966.8352\n503.3164822973.46816\n402.65318423778.77453\n322.122547224423.01962\n257.698037824938.4157\n206.158430225350.73256\n164.926744225680.58605\n131.941395325944.46884\n105.553116326155.57507\n84.4424930126324.46006\n67.5539944126459.56804\n54.0431955326567.65444\n43.2345564226654.12355\n34.5876451426723.29884\n27.6701161126778.63907\n22.1360928926822.91126\n17.7088743126858.32901\n14.1670994526886.6632\n11.3336795626909.33056\n9.06694364726927.46445\n7.25355491826941.97156\n5.80284393426953.57725\n4.64227514726962.8618\n3.71382011826970.28944\n2.97105609426976.23155\n2.37684487526980.98524\n1.9014759 ------------26984.78819\n1.52118072 -----------26987.83055\n1.216944576----------26990.26444\n0.973555661----------26992.21155\nOk, Excel columns don't work well when copied here, so I've added spaces to a few entries at the top and bottom.\n\nGuess spaces didn't work either.  Maybe the dashes will stay.
227905
When you lift a ball off the floor, you transfer energy to it. This energy is stored in the gravitational force between the ball and the earth and is called gravitational potential energy. When you release the ball, its weight makes it accelerate downward and its gravitational potential energy gradually becomes kinetic energy, the energy of motion. When the ball hits the floor, both the ball's bottom surface and the floor's upper surface begin to distort and the ball's kinetic energy becomes elastic potential energy in these two distorted surfaces. The ball accelerates upward during this process and eventually comes to a complete stop. When it does, most of the energy that was initially gravitational potential energy and later kinetic energy has become elastic potential energy in the surfaces. However, some of the original energy has been converted into thermal energy by internal frictional forces in the ball and floor. The distorted ball and floor then push apart and the ball rebounds into the air. Some or most of the elastic potential energy becomes kinetic energy in the ball, and the rising ball then converts this kinetic energy into gravitational potential energy. But the ball doesn't reach its original height because some of its original gravitational potential energy has been converted into thermal energy during the bounce.