A 10.0-g marble slides to the left with a velocity of magnitude 0.300 m/s on the
ID: 1261569 • Letter: A
Question
A 10.0-g marble slides to the left with a velocity of magnitude 0.300 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 35.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s.
(a) 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.)
m/s (smaller marble)
m/s (larger marble)
(b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble.
kg?m/s (smaller marble)
kg?m/s (larger marble)
(c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble.
J (smaller marble)
J (larger marble)
Explanation / Answer
Momentum conservation
35*0.2 - 10*0.3 = 10 Vb - 35 Va
Elastic collision
Vb + Va = 0.5
Vb = 21.5/45 = 0.4777 m/sec (Smaller marble) (ans)
Va = 0.0222 m/sec (Larger marble) (ans)
b) Change in momentum = 0.035 ( 0.2+ 0.0222) = 0.00777 Kg m/sec (ans) (Larger marble)
Chaneg in momntum of smaller marble = 0.00777 Kg m/sec (ans) (smaller marble)
c) Change in kinetic energy = 1/2*35*(0.2^2 - 0.0222^2) = 0.00069 J (ans)
Change in kinetic energy of smaller marble = 0.00069 J (ans)
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