A 1.2 kg skateboard is coasting along the pavement at a speed of 5m/s when a .8kg cat drops from a tree?
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Horizontal momentum is conserved.\n\nInitial horizontal momentum is the cats (0) + the skateboard's (1.2 kg * 5.0 m/sec) = 6 m-kg/sec\n\nfinal horizontal momentum is 2.0 kg * s, where s is the speed of the athletic cat.\n\n2s = 6\n\nSpeed s = 3.0 meters/sec
178761
A cat doesnt have feet it has paws.
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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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1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
95497
This a momentum question.\n\nRemember momentum is:\n\nm * V\n\nm = mass\nV = velocity\n\nSo the conservation of momentum is:\n\n(m1 * V1) + (m2 * V2) = (m1 * V3) + (m1 * V4)\n\nMomentum (before collision) = Momentum (after collision)\n\n(Notice mass does not change after the collision, but velocity does.)\n\nSo let's plug in the variables:\n\nm1 = 2.0 kg\nV1 = 3.5 m/s\n\nm2 = 4.5 kg\nV2 = -5.0 m/s \n(negative indicates object is traveling to the left)\n\nV3 = ?\nV4 = -1.8 m/s\n\n(2.0 * 3.5) - (4.5 * 5.0) = (2.0 * V3) - (4.5 * 1.8)\n\n(7) - (22.5) = 2(V3) - (8.1)\n\n-15.5 = 2(V3) - (8.1)\n\n-7.4 = 2(V3)\n\nV3 = -3.7 m/s \n\n(the negative sign just indicates direction with relation to the axis; in this problem, negative indicates the object is traveling to the left)\n\nSo Object A will be traveling to the left at 3.7 m/s after the collision.
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Final veloc. when he reaches the land:\nV=sqr(2*g*h)= 8.743 m/s\nKinetic Energy= Ec= 0.5 * m* V2 = 1872,780 \nThis turns into potential Energy Ep= 1872,780  = mgh\nHe stops in h=0.8m , hence: \nForce = m g = 1872,78 / 0,8 = 2340,795 Newtons\n\nHe has two knees, so F= 2340, 795 / 2 = 1170,488 Newtons in each knee.
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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
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For maybe a half a second. The ball will initially go as fast as the car, but, because the mass of the ball has so less mass in comparison to the car, it will quickly lose momentum and slow down and fall to the ground.
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Skater1 m1=66.6kg holding the snowball m3=0.162kg and running with v1=2.26m/s has momentum  M1=(m1+m3)*v1. Now he throws the ball with speed v3=36.9m/s, thus momentum of the ball M3=m3*v3. According to momentum conservation law the momentum of skater1 now is M4=M1-M3, or m1*v4=(m1+m3)*v1-m3*v3, or v4=v1+(m3/m1)*(v1-v3)=2.176m/s is the skater1’s speed. \nAccording to momentum conservation law the momentum of skater2 M2=M3, or m2*v2=m3*v3, m2=62.3kg. Thus v2=m3*v3/m2=0.096m/s – is it snowing where u live?
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Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s
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It's around 23.6 mph (penny falling flat) to 73.3 mph (penny falling on its side).\n\nVterm =  sqrt ((2 * m * g) / (c * p * A))\n \nwhere:\nm=mass of the object \ng=acceleration from gravity \nc=coefficient of drag \np=density of air \nA=Area of the object \n\nIt's pretty straightforward.  The full explanation is at the source.
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let us say that she is walking in the x direction (positive) \n\nThis is a question of momentum... \np = mv   \n(momentum equals mass times velocity)\n\nand we have the conservation of momentum \np_1  +p_2 = p_(total)\nThe sum of the momentums is equal to the total momentum of the system\n\nSo in the begining she is not moving... so the total momentum of the system is zero... it will always be zero.  \n\nShe walks... her momentum is p_1 = 45*1.5 = 67.5 kg*m/s\n\nand we know that p_1 + p_2 = 0\n\nso \np_1 = -p_2 \nor \n67.5 = - m v = - 150 *v\nso the velocity of the plank is -0.45 m/s.  \n\nNow we just sum the velocities to get the total velocity and we see that she is traveling at 1.05 m/s relitive to the ice. \n\nGood question ... i have not thought about a problem like this in a long time
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Skateboarder and his buddy bodyguard\nb
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Is the speed at which it travels.
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I think the cat would land feet first because it's heavier.plus cats' whiskers are their balance so a piece of toast shouldn't interfere with that.
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though instant the bird leaves the swing we assume that the centre of gravity does not change of the system. Therefore 52*2=153*x, where x represents the swing velocity= 0.6797m/s.\n\nnow the energy for the swing is going to be conserved.  Just a great kinetic energy with potential energy and get the height. mgh=1/2mv2\n\nCalculate h.  That is the answer.\n\nIt took me quite a long time to type all this....so I hope you will give me the best answer.