a.)what impulse acts on the ball during contact?b.)if the ball is in contact for 0.020secs,what is the average force exerted on the floor?
200765
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
474418
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.
368277
the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s
434576
The ball goes from 0 to 9/s in .15 secs\n\nSo:    9meters           1\n        ------------  x    ------                  =9 /.15 m/sec2\n           second          .15secs\n\n                                                      =60 m/sec 2
281819
Since this section is "Home-Work" and I'm here, I wouldn't tell you to do your own home work.\n\n1. From Newton's second law, a =F/m, it can be seen that when a resultant force acts on a body, it experiences acceleration.  That acceleration maybe an increase/decrease in velocity and/or a change in direction.\n\n2. (a) no acceleration -same speed, same direction.\n   (b) acceleration (negative) - change in speed.\n   (c) acceleration - change in both direction and speed.\n   (d) no acceleration - no change in speed and direction.\n   (e) acceleration - change in speed.\n\n3.  average acceleration = change in speed/ time\n    Therefore (10 mph - 210 mph)/ 10 s = -20 mph (the negative means deceleration)
515581
2. A halfback, with a mass of 95kg abd moving at 4.1m/s, on an apparent breakaway for a touchdown is tackled from behind by a cornerback whose mass is 85kg running at 5.5 m/s in the same direction. What was their mutual speed immediately after the touchdown-saving tackle?\n\n[PDF]\nLinear Momentum (Ch 9) Think-Pair-Share Linear Momentum, cont ...\nFile Format: PDF/Adobe Acrobat - View as HTML\nWhen he was tackled. by an 85-kg cornerback running at 5.5 m/s in. the same direction, what was their mutual. speed immediately after the tackle? ...\nwww.sos.siena.edu/~rfinn/phys130/P130-12\n\n\n\n3. A 9500kg boxcar traveling at 16 m/s strikes a second car. The two stck together and move at a speed of 6 m/s. What is the mass of the second car?\n\nPHYSICS - THANKSGIVING EXAM\n7. A 9500-kg boxcar traveling at 16 m/s strikes a second car. The two stick together and move off with a speed of 6 m/s. What is the mass of the second car? ...\nfaculty.coloradomtn.edu/jeschofnig/class/class_jeschof/exam.htm \n\n\n\n4. A 15g bullet strikes and becomes embedded in a 1.1 kg block of woob placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.25 and the impact drives the block a distance of 9.5m before it comes to rest, what was the muzzle speed of the bullet?\n\nhttp://www.google.com/search?hs=avo&hl=en&lr=&client=firefox-a&rls=com.google%3Aen-US%3Aofficial&q=15+g+bullet+strikes+1.1+kg+block+of+wood+placed+on+a+horizontal+surface+just+in+front+of+the+gun.+If+the+coefficient+of+kinetic+friction+between+the+block+and+the+surface+is+0.25+&btnG=Search
27115
Initially, the ball is pushed, it slides. There is no rotation, as stated in the problem/question. This means there is no rotation at all. That all the velocity is translational.\n\nFor this scenario think of the ball as a solid block that has been pushed and it sliding across the floor.\n\nAs stared, at some time (t1) the ball begins to roll.\n\nAt this time, the ball's motion is both rotational as well as translational. In other words, it rolls as well as moving forward. This motion is what takes place at the *contact point* of the ball to the floor.\n\nIf the ball rolls, it is not slipping.\n\nIndeed, the friction causes the linear motion to decrease, and eventually stop the ball.\n\nThe rotational motion of the ball is also undergoing friction. A rolling sphere's motion is very subtle and often misunderstood. By virtue of the ball's mass there is a very tiny (tiny is relative to the mass and material) flat spot between the ball and the surface upon which it rolls, whereby a small torque gives rise to a resistance of the motion. This is why a rolling sphere slows. But.....since there is friction, the velocity of translation, is related to that of rotation by the following:\n\nv^2 = 2gh/(1 + I/Mr^2)\n\nWhere M is the mass of the ball and "r" its radius, I its moment of inertia. This is common notation. If there is no rotation, then the moment of inertia term drops out (I = 0) and you are left with a velocity which is strictly that of a translating body......kind of like free-fall.\n\nIf the ball is given one initial push to give it the initial velocity Vo, then unless there is some other forcing function at work, friction will act to decrease the motion of the ball, and all said velocities thereof. None of the velocities will increase.\n\nIf you know of a situation where such a case can exist where the ball increases velocity without doing anything to it.....make sure you mention my name when you pick up the your Nobel Prize.\n\n:)
494173
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.
439685
Equation:\n\nV^2 = U^2 + 2 A (delX)\n\nThen solve F = M A\n\nwhere V = Final Velocity\nU = Initial Velocity\nA = Acceleration (unknown)\ndelX = distance of applied force (delX meaning delta X, or change in distance)\nF = Force\nM = Mass\n\nOf course you have to consider two things:\nThe force applied over the two feet would have to be constant (this is an assumption you have to make to solve)\nAll your units are in ft and weight.  you have to use the meters measurements and convert your ball's weight of 0.25 lb to kilograms (that's the mass measurement)\n\nyou now have more than enough information to solve this problem on your own.\n\nNOTE: Jim, that is a horrible misuse of the wrong equations for solving this kind of problem.  Nice try though.
279221
C, gravity is an acceleration, it is by definition not uniform then.  \n\nit's 9.8 meters per second per second, just in case you wanted to know, that's also written 9.8 m/s^2\n\nhere's a little more: it's not A, (see above), it's not B, since again this would mean that it's uniform, since 10 seconds is twice as great as 5 seconds, and it's not D because it would fall faster in a vacuum than in air since there would be no resistance.  if you drop a pen and a piece  of paper right now, the pen should fall faster since the air will resist the paper more, but in a vacuum both would hit the ground at the same time.
364210
initial vertical velocity = 30 sin 36.0 degrees = 17.6 m/s\n\nv = u + at\n0 = 17.6 + (-9.81)t\nt = 1.794 s\n\ntime that the ball is in the air = 1.794*2 = 3.588s (since it takes the same time for the ball to rise up to max height and for the ball to fall from the ground to max height )\n\n-----------------------------------------------------------------------------------\ns = ut + 1/2at^2\ns = (17.6)(1.794) + (0.5)(-9.81)(1.794^2)\ns = 15.85 m\n\nthe ball rises to 15.85 m\n-----------------------------------------------------------------------------------\n\nhorizontal speed = 30 cos 36.0 = 24.27 m/s\n\nhorizontal distance = 24.27(3.588) = 87.08 m\n\n------------------------------------------------------------------------------------\n\nplease do redo the calculations yourself cuz i didn't take much care of significant figures and i may have made errors when pressing the calculator
348740
(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)
213342
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
465098
You don't need the mass of the ball. That isn't going to change, so just pick a mass and stick with it. You won't be able to solve for momentum, but you can solve for *specific* momentum -- that is, momentum per unit mass of the ball.
539898
Apply v^2 = u^2 + 2as vertically;\nwhere v=end velocity, u=start velocity, a=acceleration, s=distance\nWhen the object reaches the height of its path, vertical velocity is zero.\nhence: 0=(30Sin30)^2 + 2(-10).s\ns = 225/20 = 11.25m. \nHence it reaches a total height of 11.25m+1.2(height from which it is thrown).\n.\n.\nTime of flight:\nApply: S=ut + 1/2 at^2 vertically, where S=distance, u = start velocity, a=acceleration, t=time\nSince the ball is caught at the same height at which it is thrown, the overall vertical distance travelled is zero.\n0 = 30Sin30.t + 1/2(-10).t^2\nt = 3s.\n.\n.\nHorizontal distance = horizontal velocity x time\ndistance = 30Cos30 x 3= 77.9 metres
657797
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..
583712
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1