A 2.0 g particle moving at 5.0 m/smakes a perfectly elastic head-on collision wi

Question

A 2.0 g particle moving at 5.0 m/smakes a perfectly elastic head-on collision with a resting 1.0 gobject. (a) Find the speed of each after thecollision. 2.0 g particle 1 m/s 1.0 g particle 2 m/s (b) Find the speed of each particle after thecollision if the stationary particle has a mass of 10 g. 2.0 g particle 3 m/s 1.0 g particle 4 m/s (c) Find the final kinetic energy of theincident 2.0 g particle in the situations described in (a) and(b). KE in part (a) 5 J KE in part (b) 6 J
A 2.0 g particle moving at 5.0 m/smakes a perfectly elastic head-on collision with a resting 1.0 gobject. (a) Find the speed of each after thecollision. 2.0 g particle 1 m/s 1.0 g particle 2 m/s 2.0 g particle 1 m/s 1.0 g particle 2 m/s (b) Find the speed of each particle after thecollision if the stationary particle has a mass of 10 g. 2.0 g particle 3 m/s 1.0 g particle 4 m/s 2.0 g particle 3 m/s 1.0 g particle 4 m/s (c) Find the final kinetic energy of theincident 2.0 g particle in the situations described in (a) and(b). KE in part (a) 5 J KE in part (b) 6 J
KE in part (a) 5 J KE in part (b) 6 J
2.0 g particle 1 m/s 1.0 g particle 2 m/s

Explanation / Answer

(a)   m1 = 2 g = 0.002 kg         m2 = 1 g =0.001 kg          U1 = 5m/s          U2 = 0m/s       Final velocity of if m1 ,V1 = [ ( m1 - m2 ) / ( m1 + m2 ) ] U1 = ( 0.001 / 0.003 ) 5 =1.67 m/s       Final velocity of m2 , V2= [ ( 2 m1 / ( m1 + m2 ) ] U1 = ( 0.004 / 0.003 ) 5 = 6.67m/s (b)   m1 = 2 g = 0.002 kg         m2 = 10 g =0.01 kg          U1 = 5m/s          U2 = 0m/s       Final velocity of if m1 ,V1 = [ ( m1 - m2 ) / ( m1 + m2 ) ] U1 = ( 0.001 / 0.012 ) 5= 0.4167 m/s       Final velocity of m2 , V2= [ ( 2 m1 / ( m1 + m2 ) ] U1 = ( 0.004 / 0.012 ) 5 = 1.67m/s (c)   Kinetic energy forcase(a):         Final K.E. of the first particle, KE1 = (1/2) m1 (V1) 2 = 0.5 * 0.002 * 1.672 =0.00279 J    Kinetic energy for case(b):    Final K.E. of the first particle, KE1 = (1/2) m1 (V1) 2 = 0.5 * 0.002 * 0.4167 2= 0.000174 J                m2 = 10 g =0.01 kg          U1 = 5m/s          U2 = 0m/s       Final velocity of if m1 ,V1 = [ ( m1 - m2 ) / ( m1 + m2 ) ] U1 = ( 0.001 / 0.012 ) 5= 0.4167 m/s       Final velocity of m2 , V2= [ ( 2 m1 / ( m1 + m2 ) ] U1 = ( 0.004 / 0.012 ) 5 = 1.67m/s (c)   Kinetic energy forcase(a):         Final K.E. of the first particle, KE1 = (1/2) m1 (V1) 2 = 0.5 * 0.002 * 1.672 =0.00279 J    Kinetic energy for case(b):    Final K.E. of the first particle, KE1 = (1/2) m1 (V1) 2 = 0.5 * 0.002 * 1.672 =0.00279 J    Kinetic energy for case(b):    Final K.E. of the first particle, KE1 = (1/2) m1 (V1) 2 = 0.5 * 0.002 * 0.4167 2= 0.000174 J       

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