A 2.0-g particle moving at 5.6 m/s makes a perfectly elastic head-on collision w

Question

A 2.0-g particle moving at 5.6 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. 2.0 g particle 1.0 g particle m/s m/s (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. 2.0 g particle 10.0 g particle m/s m/s (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b) KE in part (a) KE in part (b) In which case does the incident particle lose more kinetic energy? case (a) case (b)

Explanation / Answer

a)

m1 = 2 g = 0.002 kg

m2 = 1 g = 0.001 kg

v1i = velocity of m1 just before collision = 5.6 m/s

v2i = velocity of m2 just before collision = 0 m/s

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(0.002) (5.6) + (0.001) (0) = (0.002) v1f + (0.001) v2f

v2f = (0.0112 - (0.002) v1f ) /(0.001) eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(0.002) (5.6)2 + (0.001) (0)2 = (0.002) v21f + (0.001) ((0.0112 - (0.002) v1f ) /(0.001))2

v1f = 1.86 m/s

using eq-1

v2f = (0.0112 - (0.002) v1f ) /(0.001) =  (0.0112 - (0.002) (1.86)) /(0.001) = 7.5 m/s

b)

m1 = 2 g = 0.002 kg

m2 = 1 g = 0.01 kg

v1i = velocity of m1 just before collision = 5.6 m/s

v2i = velocity of m2 just before collision = 0 m/s

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(0.002) (5.6) + (0.01) (0) = (0.002) v1f + (0.01) v2f

v2f = (0.0112 - (0.002) v1f ) /(0.01) eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(0.002) (5.6)2 + (0.01) (0)2 = (0.002) v21f + (0.01) ((0.0112 - (0.002) v1f ) /(0.01))2

v1f = - 3.73 m/s

using eq-1

v2f = (0.0112 - (0.002) v1f ) /(0.001) =  (0.0112 - (0.002) (-3.73)) /(0.01) =1.86 m/s

c)

in part a)

KE2f = (0.5) m2 v2f2 = (0.5) (0.001) (7.5)2 = 0.028125 J

In part B)

KE2f = (0.5) m2 v2f2 = (0.5) (0.01) (1.86)2 = 0.017298 J

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