A 10.0-g marble slides to the left with a velocity of magnitude 0.350 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.350 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 20.0-g marble sliding to the right with a velocity of magnitude 0.250 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

larger marble

initial velocity = v1i = +0.25

smaller marble

initial velocity = v2i = -0.35

from momentum conservation

P1i + P2i = P1f + P2f

m1*v1i + m2v2i = m1v1f + m2v2f

m1*(v1i-v1f) = m2*(v2f-v2i)

from energy conservation

0.5*m1*v1i^2 + 0.5*m2*v2i^2 + 0.5*m1*v1f^2 + 0.5*m2*v2f^2

m1*(v1i^2-v1if^2) = m2*(v2f^2-v2i^2)

v1f = (((m1-m2)*v1i) + (2*m2*v2i))/(m1+m2)

v2f = (((m2-m1)*v2i) + (2*m1*v1i))/(m1+m2)

(a)

v1f = (((20-10)*0.25)-(2*10*0.35))/(20+10) = -1.5 m/s

v2f = (((10-20)*-0.35)+(2*20*0.25))/(20+10) = 0.45 m/s

(b)

P2f - p2i = m2*(v2f-v2i) = 0.01*(0.45+.35) = 0.008 kg m /s

P1f - p1i = m1*(v1f-v1i) = 0.02*(-1.5-0.25) = -0.035 kgm/s

c)

dKE2 = 0.5*m2*(v2f^2-v2i^2) = 0.5*0.01*(0.45^2-0.35^2) = 0.0004 J

dKE1 = 0.5*m1*(v1f^2-v1i^2) = 0.5*0.02*(1.5^2-0.25^2) = 0.0219 J

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