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 0.150 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

Mass of the small marble, m = 10 x 10 -3 Kg = 10-2Kg = 0.01 Kg

velocity of the small marble, v1 = 0.350 m/s

Mass of large marble, M = 20 gm =20 x 10 -3 Kg = 2x10-2= 0.02 Kg

velocity of the large marble, V1 = 0.150 m/s

Since the collision is elastic, we can have V1 - v1 = V2 - v2

Let  v1 and V1 are the velocities before collision along x-axis and  v2 and V2 are the velocities after

collision, respectively.

A) According to law of conservation of momentum

MV1 + mv1 = MV2+mv2

= 0.02 x 0.150 + 0.01 x (-0.350) =  0.02xV2 + 0.01 x v2 --------------(1)

2V2+v2 = -0.0005-----------(1)

But from the equation, V2 - v2= -(V1 - v1)

= -(-0.350-0.150)

= 0.500 m/s   

Hence, V2 - v2 =0.500 m/s

So, V2 = v2 +0.500------------(2)

Substituting the above value in equation (1), we get

Solving (1) and (2), we get

2(v2 +0.500)+v2 = -0.0005

3v2 = -1-0.0005 = -1.0005 m/s

So, v2 = -0.3335 m/s

Since, V2 = -0.3335 +0.500

= 0.1665 m/s

(b) The change in momentum for small marble = 1/2 x0.01x(0.3335+0.350) = 0.00342 Kg m/s

The change in momentum in large marble = 1/2 x 0.02 x (-0.1665-0.150) = 0.01 x -0.3165 =-0.003165 Kgm/s

c) The change in K.E. for small marble, m = 1/2m(v22- v12)

= 1/2 x 0.01 x [(-0.3335)2 - (0.350)2] -5.13 x 10-5J

The change in K.E. for large marble = 1/2m(V22- V12) = 1/2 x 0.02 x [(0.1665)2 - (0.150)2] -5.22 x 10-5J

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