A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the
ID: 2259599 • Letter: A
Question
A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 25.0-g marble sliding to the right with a velocity of magnitude 0.150 m/s.
A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 25.0-g marble sliding to the right with a velocity of magnitude 0.150 m/s. Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. Take right as the positive x direction.) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble.Explanation / Answer
a). from momentum conservation; 10*10^-3*(-0.5) + 25*10^-3* 0.15 = 10*10^-3* v1 + 25*10^-3 *v2 => -5 + 3.75 = 10*v1 + 25*v2 => v1 = -(1.25 +25*v2)/10 from energy conservation (since the collision is elastic); 0.5*10*10^-3* (0.5)^2 +0.5*25*10^-3 *0.15^2 = 0.5*10*10^-3*v1^2 + 0.5*25*10^-3*v2^2 => 2.5 + 0.5625 = 10*v1^2 + 25*v2^2 => 3.0625 = 10*[-(1.25 +25*v2)/10]^2 + 25*v2^2 => v2 = -0.2214 m/s v1 = 0.4285 m/s b). change in momentum of smaller marble = 10*10^-3*[0.4285- (-0.5)] = 9.285*10^-3 kg-m/s change in momentum of bigger marble = 25*10^-3*[-0.2214 -0.15] = -9.285*10^-3 kg-m/s c). change in K.E. of small marble = 0.5*10*10^-3 *[(0.4285)^2 -(-0.5)^2] = -3.3163* 10^-4 J change in K.E. of small marble = 0.5*25*10^-3 *[(-0.2214)^2 -(0.15)^2] = -3.3163* 10^-4 J
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