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.
0.008 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.
0.004 J (smaller marble)
_____ J (larger marble)

Explanation / Answer

here,

Let,
Vai = Velcoity of 10g mbrble = 0.350 m/s
Ma = Velcoity of 10g marble = 10g = 0.01 kg

Vbi = Velcoity of 20g marble = 0.250 m/s
Mb = Velcoity of 20g marble = 20g = 0.02 kg

part A: for head on collision

Vaf = ( (Ma-Mb) / (Ma+Mb) )*Vai

Vaf = ( (0.01 - 0.02) / (0.01 + 0.02) ) * 0.350

Vaf = -0.116 m/s

and

Vbf = [(2Mb)/(Ma+Mb)]Vai

Vbf = ( 2*0.02 / (0.01 + 0.02) ) * 0.350

Vbf = 0.466 m/s

Part B:
from conservation of momentum we have :

Pa = MaVaf - MaVai = Ma(Vaf-Vai)
Pa = 0.01(-0.166 - .350)
Pa = -0.0051 Kg.m/s

Pb = MbVbf - MbVbi = Mb(Vbf-Vbi)
Pb = .02(0.466 - 0.250)
Pb = 0.00432 Kg.m/s

Part C:
changein kinetic Energy = (K.E)f - (K.E)i

0.5*maVf^2 - 0.5*maVf^2
Change in kinectic energy for 10g ball = 0.5*Ma*(Vaf-Vai)^2 = 0.5*.01*(-.116 - 0.350)^2

Change in kinectic energy for 10g ball = 0.00108 J

Similarly,
Change in kinectic energy for 20g ball = 0.5*Mb*(Vbf-Vbi)^2 = 0.5*.02*(0.466- 0.250)^2

Change in kinectic energy for 20g ball = 0.00046 J

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