Question

A 2.00-9 particle moving at 7.40 m/s makes a perfectly elastic head-on collision with a resting 1.00-9 object. (Assume the 2.00-9 particle is moving in the positive direction before the collision. Indicate the direction with the sign of your answer.) (a) Find the velocity of each particle after the collision. 2.00-9 particle 1.00-9 particle m/s m/s (b) Find the velocity of each particle after the collision if the stationary particle has a mass of 10.0 g. 2.00-9 particle m/s 10.0-9 particle m/s (c) Find the final kinetic energy of the incident 2.00-9 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)

Answer 1

a)

for 2 g particle

v1 = (m1 - m2) u1/ ( m1 + m2) = (2 - 1)* 7.4 / ( 2 + 1)

v1 = 2.467 m/s

for 1 g particle

v2 = 2 m1 u1 / ( m1 + m2) = 2* 2* 7.4 / 3 = 9. 867 m/s

=====

b)

for 2 g

v1' = ( 2 - 10) * 7.4 / ( 2 + 10) = - 5.92 m/s

for 10 g

v2' = 2* 2* 7.4 / (2 + 10) = 2.467 m/s

=====

c)

kE1 = 0.5 m v1^2 = 0.0061 J

kE2 = 0.5* 2* 10^-3* 5.92^2 = 0.035 J

=====

d)

for case (a)

dkE = 0.5* 0.002* ( 7.4^2 - 2.467^2) = 0.04867 J

for case (b)

dkE = 0.5* 0.002* ( 7.4^2 - 5.92^2) = 0.0197 J

so, in case (a) we observe more change in kE than case(b)

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