A 10.0-g marble slides to the left with a velocity of magnitude 0.550 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.550 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 0.150 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 positivex 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

m1 =10g, u1 = -0.55 m/s, m2 =35g, u2 =0.15 m/s

(a) From conservation of momentum

m1u1+m2u2 =m1v1+m2v2

(10*-0.55)+(35*0.15) =10v1+35v2

- 0.25= 10v1+35 v2 ..(1)

In elastic collision

u1 -u2 =v2-v1

-0.55-0.15 =v2-v1

v2-v1 = - 0.7 ..(2)

By solving (1) and (2) we get

Velocity of smaller marble v1 = 0.54 m/s

Velocity of larger marble v2 = -0.16 m/s

(b) p1 =m1(v1-u1) = 10(0.54+0.55) = 10.9 m/s

Change in moment of smaller marble p1 =10.9 m/s

Change in moment of larger marble p2 =-p1 = -10.9 m/s

(c) change in kinetic energy of smaller marble = Kf1 -Ki1

= (1/2)m1v1^2 -(1/2)m1u1^2

= (1/2)(10) [(0.54)2 -(-0.55)2]

change in kinetic energy of smaller marble = -0.0545 J

change in kinetic energy of larger marble = 0.0545 J

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