document.write( "Question 1012422: A tractor of mass 5*10^3kg is used to Low a car of mass 2.5*10^3kg. The tractor moved with a speed of 3.0m/sec just before the towing rope becomes taut. Calculate
\n" ); document.write( "(1) speed of the tractor immediately the rope become taut.
\n" ); document.write( "(2) loss in K. E of the system just after the car has started moving.
\n" ); document.write( "(3) impulse in the rope when it jerks the car into motion.
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Algebra.Com's Answer #628357 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Conservation of momentum.
\n" ); document.write( "\"m%5B1%5Dv%5B1%5D=%28m%5B1%5D%2Bm%5B2%5D%29v%5B2%5D\"
\n" ); document.write( "\"5000%283%29=%285000%2B2500%29v%5B2%5D\"
\n" ); document.write( "\"15000=7500v%5B2%5D\"
\n" ); document.write( "\"v%5B2%5D=2\"\"m%2Fs\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "Initially,
\n" ); document.write( "\"E%5B1%5D=%281%2F2%29%285000%29%283%29%5E2\"
\n" ); document.write( "\"E%5B1%5D=22500\"\"J\"
\n" ); document.write( "Then,
\n" ); document.write( "\"E%5B2%5D=%281%2F2%29%287500%29%282%29%5E2\"
\n" ); document.write( "\"E%5B2%5D=15000\"\"J\"
\n" ); document.write( "So,
\n" ); document.write( "\"DELTA%2AE=E%5B2%5D-E%5B1%5D\"
\n" ); document.write( "\"DELTA%2AE=15000-22500\"
\n" ); document.write( "\"DELTA%2AE=-7500\"\"J\"
\n" ); document.write( ".
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\n" ); document.write( "Impulse equals the momentum change of the car.
\n" ); document.write( "\"I=m%28v%5B2%5D-0%29\"
\n" ); document.write( "\"I=2500%282%29\"
\n" ); document.write( "\"I=5000\"\"%28kg%2Am%29%2Fs\"
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