investigate the relationship between the temperature of a rubber squash ball and the height to which it bounce
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The previous idea of doing an experiment or two is a good idea. Here are some more ideas:\n\nThe ratio of the height you drop the ball from to its rebound height is called its coefficient of restitution. Different types of squash balls have different coefficients of restitution. \n\nThe square root of the coefficient is also the ratio of the maximum speed of descent to maximum speed of rebound (converting potential energy into kinetic energy, and vice versa).
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It's the density of the rubber that matters, not the mass of rubber, and there are rubber formulations that are less dense than water.  The easy answer is do the experiment and see.\n\nYou might always want to try it with different temperatures of water inside and outside of the balloon.
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The miximum height reached by the ball is when the velocity of the ball is zero.  Since the initial velocity is 64ft/s, the distance above the ground equation becomes s = -16t^2 + 64t + 0\ns = -16t^2 + 64t .  Taking derivaties to find velocity equation.\nds/dt = -32t +64.  At maximum height ds/dt = 0.  \nThus, -32t + 64 = 0\n            32t = 64\n              t = 2\nSo, maximum height is reached at t = 2s.  So the maximum height reached by the ball:\ns = -16(2)^2 + 64*2\ns = -64 +128\ns = 64 feet.\nThe maximum height reached by the ball is 64 feet.
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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
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They had that on an episode of Mythbusters.  The ball travelled less further because of the decrease in mass and no change in volume.  It was more susceptible to air resistance.
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Ideal gas law PV=nRT, which means Pressure x Volume = constant x Substance x Temperature.  The n remains constant and the Volume increase is so small that we'll say it is constant too.  That leaves us with P = RT, we know the R (substance) goes up since you are filling the tire with air, but we have two variables left, and depending on certain conditions the temperature or pressure can go up or down.  For that you would need to apply the second law of thermodynamics (entropy, simply stated if you put something hot next to something cold, the cold thing will get hotter and the hot thing will get colder, not vise versa) If you use hot air to fill the tire and the rubber from the tire is cold, the temp will go down (which will also decrease the pressure), if the air is colder than the tire the temp will increase (and so will the pressure).
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45 degrees.  It helps you calculate the distance and strenght you need to make the shot and if your ball bounces off the board, it has a better chance of still getting in.
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mass, volume and temperature.
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a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
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If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
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(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)
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It should be good for all conditions. Obviously the ball will fly farther when warmer due to the viscosity of rubber. \n\nBut the driver itself should perform the same no matter what the weather.
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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
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the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s
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There is a little trick to this question, Alyssia. \nWe can try to use the ideal gas law to calculate the change in volume.\n P1V1 / T1 = P2V2 / T2  \n\nKeep in mind that temperature is absolute, not celsius, so T1 is 273.15 + 20 = 293.15 and T2 is 273.15 - 196 = 77.15\n\nSolving for the second volume, we get the equation V2 = V1 * P1/P2 * T2/T1.  \nSubstituting the given values we have V2 = 2.91 * 1/5.2 *  77.15/293.15\nTherefor the new volume might be .147 L.\n\nHowever, the balloon will loose all elasticity before reaching that temperature.  Therefor the actual starting and ending pressures of the gas INSIDE the ballon cannot truely have the same ratio as the starting and ending pressures outside the ballon.  For a real balloon, the answer is too difficult for me to calculate.  \n\nFor a theoretical balloon that keeps a constant elasticity, the answer is .147 L.
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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.
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Simple parabolic path formula \n\na hit ball is given a velocity v at and angle theta.\n\nAssuming no air resistance the balla travels through the air along its horizontal vector un abstructed, while acted on by gravity in the vertical direction.\n\nthe equation\n\nd=vt + 1/2gt^2 can be solved for d which will be the total vertical distance at time t, where t is the time at which the ball is the highest, lets say about 1/2 of the total time, g is the gravitational constant and v in the velocity in the vertical direction