A 900 g air-track glider collides with a spring at one end of the track. How long is the glider in contact wit
480504
first, you would need to know the final velocity of the glider (velocity right when it touches the spring), secondly, you would need to know the spring constant, k.  obviously, the greater the spring constant, the shorter the time of contact would be.  you would then have to go through various steps:\n\nin a nuttshell:\n1)  find the force/momentum of the glider when it hits\n2)  from the info in #1, you will be able to find the length of the springs compression, as well as the time it takes for it to bring the glider to rest.\n3)  then using the spring constant, you'd need to find out how long it'd take for the spring to push a glider (remember, the glider is at rest, ie. v-knot=0) of 900g from its compressed state (found in step 2).\n4)  adding the times found in steps 2 and 3 should give you your final answer.
957506
k = .7/.25 = 2.8 N/m\n\nT = 2π√(m/k) = 2π√(.25/2.8) = 1.877 sec\n\nergo, t = .71*1.877 = 1.33267 sec\n\nYou haven't told us where the glider is at t = 0\nIf it's at its deflected position, then x = .25cos(2πt/T) = -.062\nIf it's at x = 0 moving left, then x = -.25sin(2πt/T) = +.218
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.
447816
the track is 400 meters around (1 lap).\nFor every 1 yard = 0.9144 meters.\n\n400 meters = 437.45 yards (1 lap around the track)
1144696
gravity and air resistance.
332219
An egg is pretty aerodynamic, so we can neglect friction. Thus, a falling object satisfies the equation\n\ns(t) = -0.5*g*t^2+v0*t+s0\n\nwhere g is 9.82 m/s^2, v0 is the start velocity (in this case, 0) and s0 is the start height (in this case, 3 metres).\n\nSo, letting s(t)=0 and solving for t, we get\n\n0 = -0.5*g*t^2 + 3\n\nt = sqrt(6/g) = 0.78 seconds.
698592
The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
152421
Well I wont solve it for you \nbut I will help you. \n\nThese problems always involve the equation \ns=s0+v*t+(at^2)/2\n\nwhere s is displacement\nv is speed\nand a is acceleration\n\nNow you will always have acceleration of gravity g\n\nbut when the freak comes out of the canon, they will have a certain velocity. \n\nThey will be accelerated as the spring releases its energy, and then leave the barrel with a constant velocity, at that angle. \n\nThey will have a horizontal velocity and a vertical velocity. \n\nso you will need two equations\n\nx= Vx *t + .5*Ax*t^2\n\nand y=Vy*t+.5*Ay*t^2\n\nOne for vertical travel, and one for horizontal travel\n\nThe horizontal travel will will assume Ax=0 (not really true, but close enough)\n\nThe vertical travel, assume Ay=-g, acceleration of gravity. \n\nTo find the speed out of the barrel at the start, \n\nremember Hooke's law\n\nF=kx. \n\nSo you know k and you know x, so you know F\n\nYou also know Newton's 2nd law, F=ma.\n\nSo you know a =F/m\n\nNow you have to find the velocity V. \n\nAs the spring releases, it accelerates the freak with acceleration a. The only question is, over what distance is this acceleration acting? \n\nWell one idea is that it acts over the compressed distance. Another idea is that it acts over 2 times the compressed distance, since it will oscillate in approximately simple harmonic motion, and will continue to push that entire distance. I would guess the latter.\n\nSo put this in \n\ns=.5*at^2, \nand compute t\n\nFrom t, you can figure out the speed at launch: \n\nV=a*t\n\nSo you have the initial velocity V, which you can resolve into vertical and horizontal components, and then figure out how long it takes the freak to return back to the ground, and in that time, how far will the freak travel horizontally. \n\ngood luck\n\n- - - - - - - - - -\n\nkirchwey is correct. The same force is not applied through the entire distance that the spring oscillates. The force F=kx is the maximum force, only at the beginning of the freak's acceleration. I was thinking about that so I returned to this problem (it has been a long time since i did these problems). \n\nThe simple way to address this is to assume that all of the potential energy of the spring .5*k*x^2 is imparted to the freak, and turned into the freak's kinetic energy, .5m*v^2. \n\nAs kirchwey points out, there might very well be a new equilibrium position for the spring due to the freak's weight, or at least the component of the freak's weight that is projected along the axis of the tube. If the spring was not at all compressed by anything but the freak's weight, then if the spring was "released" the freak would not go anywhere. What one has to decide is, is the equilibrium point the problem mentions the equilibrium point for an unloaded spring, or the spring with freak compressing it a bit. Whatever you decide will determine how you solve this problem. I would try it both ways and see what the different answers are (they should not be too different because the freak does not weigh that much, relative to the spring force).
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps
812983
I solved this as if the artist strikes the trampoline at 5 m/s downward.\nIs he standing on the trampoline and gets sent upward at 5 m/s, or is he standing on a platform some distance above the trampoline?\n\nFor striking the trampoline at 5m/s downward\nStart by considering the energy in the system:\nThe artist has kinetic energy of 1/2*m*v^2\nand potential energy of m*g*d\nwhere g is 9.8 m/s^2 and d is the net downward displacement of the spring\n\nThe spring will absorb the energy =1/2*k*d^2\nwhere k is the spring constant\nat equilibrium when the spring is fully compressed the energy in the spring will balance out the rest\nset all energy equal to zero for equilibrium\n0=1/2*kd^2-m*g*d-1/2*m*v^2\nmultiply by two\n0=k*d^2-2*m*g*d-m*v^2\n\nplug in the numbers\n0=6.2E+4*d^2-2*65*9.8*d-65*5^2\n0=6.2E+4*d^2-1274d-1625\nthe positive root, which is the displacement of the spring, \nset negative since I used positive motion downward\nd=-.172m\n\nSince this isn't the answer, I will assume that the artist was on a platform and jumped onto the trampoline.  I will compute the height of the platform above the uncompressed trampoline:\nIn this case the spring energy is computed from your answer:\n1/2(62000)*.31^2\n=2979\nthe potential energy lost due to compression is\n65*9.8*.31\n=197.5\nHis vertical velocity when he jumped from the platform\n=.5*65*25\n=812.5\nThe energy will sum to zero.\nThe energy in the spring came from three sources:\nHis vertical velocity in kinetic energy\nPotential energy by standing on a platform above the trampoline\nThe potential energy given to the spring due to his weight and travel during compression\nSo\n2979-197.5-812.5-65*9.8*h=0\nwhere h is the height of the platform\nh=1969/637\nh=3.1m\n\nj
1060398
Two airplanes are both travelling along linear courses, at fixed speeds.  Both will pass over a specific place.  At the same time, one plane is 500 miles east of the place; the other plane is 300 miles south of the place.  If the east plane travels at 500 mph, what speed must the other plane travel at to collide with the first?\n\n--keep in mind I've created a 3D system here--\nSet the point of collision at the origin (0,0). Now we have:\nx=500-500t\ny=300-nt.\nWe're looking for n.\nWe know that t=t in order for them to collide. \nNow solve for x=0 to find out when (the value of t) the first plane reaches the place of collision.  t=1.\nPlace y=0 and t=1 to solve for n.\nn=300mph.\n\nYou have to make your own graph.
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
153350
This problem employs the simple concept of conservation of energy (athough there are a number of ways to solve it).\n\nBasically, the person in the wheel chair has an initial kinetic energy at the bottom of thr ramp,\nKE = 1/2 mv^2\nAt the top of the ramp, some of that kinetic energy has been transfered into Gravitational Potential Energy,\nPE = mgh\n\nSince the mass of the person is constant, that drops out of the equation.\n\nSolve for the change in PE and subtract this from the intial KE.  Using the new KE, solve for the final velocity.
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
919976
Apparently we are ignoring any pitching motion due to their getting in the car and due to the bump, since you'd need to know a lot more about mass distribution if it was pitching. So we're only considering vertical movement and the distance between axles is irrelevant. I'm assuming "a total of 8.9 cm" means the sum of the 4 spring deflections, so their average deflection is 2.225 cm. If it actually means the average deflection, then k and m are 1/4 what you'll get with the following calculations. Stiffness k (of the 4 springs in parallel) in n/M = 146*g/.02225. Natural radian frequency w = sqrt(k/m), so m = k/w^2 = k/(2*pi/1.7)^2. (That's the total m; mass of car = m-146.) \nThis was not a well-phrased question, because "a total of 8.9 cm" also could mean 8.9 cm from their no-load length (and thus due to the weight of the car and the students). However this would result in an insoluble problem. \nBTW, why are they geeky? Did they figure all this out before me?