Between a moving pendulum and a moving block.\n\nI just want to know how to go through the steps in finding the maximum vertical distance after the collision...so please, no bashing :)! I didn't include the specific values.\n\nGiven that the pendulum was at rest 10 degrees below horizontal with a mass M1 and a rod length L (which is massless), and the block with a mass M2 and an initial velocity Vo2 towards the pendulum. The pendulum is released, and when it is straight vertical, the two objects collided (they stick together and travel as a composite mass)...\n\nI know I have to use Conservation of Energy, and Cons of Momentum, but since there's a vector quantity because of pendulum, I don't know what to do! Help!
207014
Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].
645938
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!
448562
good start.\n\ntotal inital momentum = 480-397.5\n\nthis is an inelasitc collision, and aferwards you have \nboth masses travelling at velocity v:\n\n(75+120)v = 480-397.5\n\nsolve for v.  This is the velocity at which they both slide.\nThe inital energy now is 1/2(75+120)v^2\n\nThis energy is converted to frictional energy:\n\ninital kinetic energy = total work of friction = Fr(distance)\n\nFr = uN = u(120+75)g = .3(195)(9.8)\n\nso we have\n\n.3(195)(9.8)(distance slided) = 1/2(195)v^2 \n\nYou know v form above.  So find distance slided.
475318
'Large' and 'small' probably refer to the masses of the balls, rather than the sizes. Assume one ball (with mass m1) is initially directly above the other (mass m2), and both are dropped at the same time from rest (with a slight vertical space between the two).\n\nThe lower ball hits first, bounces up and collides with the upper ball near the ground. The two  have about the same speed v = sqrt(2gH) going into this collision because both balls have fallen from approximately the same height. If the two masses were equal, then both would simply reverse direction, the top ball going upward at speed v and the bottom ball going down (for the second time) and immediately up at the same speed v --> both balls would continue up at the same speeds to the original height H.\n\nNow suppose the masses are slightly different with m2 > m1. Then the initial and final velocities (taking the upward direction as positive) for the collision are respectively:\n\nv1 = -v\nv2 = v\n\nand:\n\nv1 = v + x1\nv2 = -v + x2 \n\nwhere x1,x2 << v. Energy and momentum conservation require that:\n\n(m2-m1)v = m1(v + x1) + m2(-v + x2)\n(m1+m2)v^2 = m1(v + x1)^2 + m2(-v + x2)^2 \n= m1(v^2 + 2vx1) + m2(v^2 - 2vx2) (Approximately)\n-->\n2(m2-m1)v = m1x1 + m2x2\n0 = 2m1vx1 - 2m2vx2\n-->\nx1 = (m2/m1 - 1)v\nx2 = (1 - m1/m2)v\n\nSo the top (lighter) ball ends up with a higher speed (m2/m1)v and the lower (heavier) ball ends up with the lower speed (m1/m2)v. This is reflected by the one bouncing higher than H and the other lower than H.\n \nThere can't be more than one collision since the top ball comes out of the first (and only) collision with a higher speed than the bottom ball so the latter can never catch up to it. If the mass ratio were reversed (heavier ball on top) the situation would be much more complicated since there could be multiple collisions.
535156
hang on.\nok first to find the maximum x-coordinate. For this you only need to consider the 'i' parts of the vectors. This is when the velocity is zero, so you now know these:\ninitial velocity, u=4m/s    ;    \nfinal velocity, v=0m/s    ;    \nacceleration, a= -2m/s^2     \nand you want to find distance moved, s\nUse the equation of motion:\nv^2 = u^2 + 2as\n==> 0^2 = 4^2 + 2(-2)s\n==> 0 = 16 - 4s\n==> 4s = 16\n==> s = 4.\n\nSo the maximum x-coordinate is 4.\nIt's horizontal velocity is 0, so you can find it's j-velocity by first finding the time it takes to do this:\nv = u + at\n0 = 4 + (-2)t\n2t = 4\nt = 2s\n\nSo now you can find the j-velocity, \nu = 0\na = 0.5 m/s^2\nt = 2s\nand you want to find v.\nSo, you use v = u +at\nv = 0 + 0.5 x 2\nv= 1 m/s\n\na) Therefore the velocity at this point is 1 j m/s.\n\nFor the position vector you need to find the y-coordinate:\nyou want to find s:\nso, you use s = ut + (1/2) x a x t^2\ns = 0 + 0.5 x 2^2 = 0.5 x 4\ns = 2m.\nSo the y-coordinate is 2 and the x-coordinate is 4\n\nb) Therefore the position vector is (4 i  +  2 j) m
446702
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
43880
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!?
410990
a. System velocity after arachnocrash (conservation of momentum): v2 = v1*m1/m2, m1=0.05, m2=4.05, v1=3\nAngular rate = v2/r, r=0.25\nb. KE2/KE1 = m2*v2^2/(m1*v1^2)\nc. System will be stable with spider at bottom (90 deg from initial position). In the absence of friction and damping it will oscillate forever like a pendulum from its initial position through the stable position to a position opposite the initial position. With enough friction it won't move, or will move only partway toward stable position. Of course, the figure "as shown below" isn't there, so the stable position could be different. Not enough information, in that case.
434715
We can use trigonometry to determine the vertical and horizontal components of the stone's velocity.  This will let us determine how far it goes.\n\nImagine a right triangle with a hypotenuse of 14 units and one angle of 26 degrees.  The side opposite the 26 degree angle will be as long as the stone's vertical velocity.  The side adjacent to the 26 degree angle will be as long as the stone's horizontal velocity.\n\nSince:\n\nsinA = opposite / hypotenuse... \nsin26 = vertical velocity / 14...\nvertical velocity = 6.14m/s\n\ncosA = adjacent / hypotenuse;\ncos26 = horizontal velocity / 14...\nhorizontal velocity = 12.56m/s\n\nIf the object had an initial vertical velocity of 6.14m/s, we know how long it stayed in the air because:\nacceleration = change in velocity / time\nWhich is the same as saying:\ntime = change in velocity / acceleration\nThis applies in any situation where the acceleration is constant.\nSo, if we want to know how long it took for the stone to reach 0m/s vertical velocity (the point at which it stops rising and starts falling), the change in velocity will be 6.14m/s.  If we just divide this number by acceleration due to Earth's gravity (9.81m/s^2), we get 0.63 seconds.  The stone traveled upward for 0.63 seconds.  This means that it must have also traveled downward for 0.63 seconds before hitting the ground, for a total flight time of 1.26 seconds.\n\nNow, we have a simple velocity equation.  Since we know that:\ndistance = velocity * time \nSince we're only worried about how far the stone traveled horizontally, we use the stone's horizontal velocity: 12.56m/s.  Time will be the total time that the stone was moving: 1.26sec.\n\nSo...\ndistance = 12.56 * 1.26 = 15.83 meters.\n\nHope that helped!
393735
Use mgh for the pendulum at rest at the highest point, ( in other words at one end)\n\nm = mass of the "bobble" at the end of the pendulum\n\ng = force due to gravity\n\nh = height of bobble above where it is at the centre\n\nthe mgh above is the potential energy of the bobble due to its height (it is at rest for a moment as it changes direction.\n\nAt the centre it no longer has any potential energy, as it is at its lowest point and potential energy is due to height.\n\nAll its energy at the centre is kinetic energy\n\nthis is 1/2 m v^2\n\nsince energy is neither created nor destroyed ( at least for the purposes of questions like this)\n\nmgh = 1/2mv^2\n\nDivide each side by m (that gets rid of the bobble).\n\nThe question will have told you the height (or enough about the circle for you to find it)\n\nIt's not a very clear diagram.  It looks to me as if the height is 50cm = 1/2 metre\n\ngood luck!
32330
First Question:\nVertical and horizontal motion are independent from each other.  Use d=(v.i)(t)+(1/2)(a)(t)^2 to find the time that the dart is in the air.  d is distance(0.32m), v.i is the initial velocity(0.00m/s vertically since it was thrown exactly horizontally), a is acceleration (9.8m/s^2: gravity) and t is the time in seconds.  Now that you have the time, use that in the equation x=vt, where x is the horizontal distance (range) that you are asked to solve for, v is the velocity(11.8m/s), and t is the time that you just found.  The answer to the problem is the x variable in this equation.\n\nSecond Question:\nI have attempted this problem for about 40 minutes and I have tried every possible combination of variables and equations, but there is not enough information to solve it.  Therefore, I have to say that it is an impossible problem.  If you figure out how to solve it, please, send me an e-mail at the address I give later and tell me how to do it.\n\nThird Question:\nFirst, find the distance that the basket is from the launch point by subtracting the two.  Then, use (v.f)^2=(v.i)^2+(2)(a)(d), where v.f is the final velocity, v.i is the initial velocity, a is the acceleration (9.8m/s^2), and d is the distance which you just found.  To  get the ball to the basket, it only needs to go the exact distance to the basket, so the final velocity is 0.00m/s.  Use the previous equation to find the intial vertical velocity.  The use v.iy=(v)sin(angle), where v.iy is the initial vertical velocity and v is the initial angular velocity, to find the velocity the ball must be launched at for each angle.\n\nPlease send me an e-mail at bge1234@yahoo.com after you turn in your work and get the solution so that I can see how your teacher did it.  In particular, tell me how to do the second question.  My ways SHOULD work as long as you don't make any calculatory errors.  I hope this helps you.
350607
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
416477
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
480617
Your description of the problem is not iron-clad, so let me make these assumptions explicit:\n- The axis of the cylinder is horizontal.\n- The block descends because of acceleration due to gravity(g).\n-  There is no air resistance to the motion of the block.\n- The string does not slip on the surface of the cylinder and it is\n  completely inelastic.\n\nAcceleration of the block is  = g\nForce acting on the block is = mg\nThe torque by this force on the cylinder about the axis = mgr\nThis results in accelerating the rotation of the cylinder.\nmgr = I x alpha, where I = moment of inertia of the solid cylinder;\n                         alpha = Angular acceleration of the cylinder.\n\nI = (1/2)mr^2 \n[ M.I of a solid cylinder; Given: Mass of cylinder = block = m ]\n\nalpha = torque/I = mgr/I = 2g/r
511220
1) The total rocket, with mass 3M travels 260m over the first "half" of the parabola. On exploding, the fragment with mass M falls straight down - i.e., all the momentum in the horizontal direction remains with the larger fragment. Therefore its speed in the horizontal direction will increase:\n\nP = constant\nBefore explosion, P = (3M)(u)\nAfter explosion, P = (2M)(v)\n\nTherefore the speed in the horizontal direction will increase to 3/2 times the original speed.\n\nThe vertical momentum of the larger fragment will not increase, as the smaller fragment falls as from rest i.e. no original momentum. Therefore the time taken for the larger fragment (indeed, either fragment) to reach the ground from the apex of the motion will be the same as it was to reach the apex from the starting point. Since the horizontal distance was 260m, the extra distance covered will be that multiplied by 3/2, since the speed increased by that much, so the total distance from the launch point is 260 + 3/2(260) = 650m.\n\nYou do it by considering the horizontal and vertical motions independently.\n\n2) If the bobs collapse elastically, then no energy is lost in the collision. The energy of the larger bob on contact is entirely its kinetic energy - this comes entirely from the potential energy it had at the top of its swing. If the angle is 15deg and the length of the cord is 50cm, then the bob is drawn back until it is at a height of 1.7cm. The formula for potential energy is U = mgh, so the potential energy of the bob at that point is \n\nU1 = (0.39kg)(9.8m/s2)(0.017m) = 0.06497 joules\n\nconverting all units to SI.\nThis is the kinetic energy of the bob as it strikes the second one - all of this energy is transferred to the second bob, as kinetic - the second bob rises until it stops, when all the kinetic energy is converted to potential - and so use the same formula to get the height it reached.\n\nU2 = 0.06497 J = (0.14kg)(9.8m/s2)(H)\n\nwhere H is the maximum height of the lighter bob. It works out as\n\nH = 0.047 m\n\nor 4.7 cm. Using simple trigonometry, this corresponds to an angle of 25.1 deg.\n\nHope this helps
394670
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
484875
So I'm assuming there is no static or kinetic friction in this problem.\n\nWell, since the cart experiences a force in the horizontal direction, then: \nF = ma\na = F/m\na = 11N/26kg = 11/26 m/s^2\n\nHow far will it move in 2 seconds starting from rest?  What do we know?\n\na = 11/26 m/s^2\nt = 2 secs\nVelocity initial = 0 m/s (since it starts from rest)\nSo what is d (for distance although I know some schools use X)?\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/26)(2)^2\nd = 11/52*4\nd = 11/13 m\n\n30 N child in the cart before he begins to push it?\n\nNewtons is just another way of saying weight but in order to find the acceleration we need mass and not weight.\nWeight (in N) = mg\nm = weight/g\nm = 30/10\nm = 3kg  <-------- this is the mass of the child\n\nRepeat as above with little modification.\na = F/m\na = 11N/(26kg + 3kg) = 11/29 m/s^2\n\nd = V(initial)t + 1/2at^2\nd = 0t + (1/2)(11/29)(2)^2\nd = 11/58*4\nd = 44/58 m\n\nGood luck!