A 2.0-g particle moving at 8.6
A 2.0-g particle moving at 8.6 m/s makes a perfectly elastichead-on collision with a resting 1.0-g object. (a) Find the speedof each particle after the collision.
| 2.0 g particle | m/s |
| 1.0 g particle | m/s |
(b) Find the speed of each particle after the collision if thestationary particle has a mass of 10 g.
| 2.0 g particle | m/s |
| 10.0 g particle | m/s |
(c) Find the final kinetic energy of the incident 2.0-g particle inthe situations described in parts (a) and (b).
| KE in part (a) | J |
| KE in part (b) | J |
In which case does the incident particle lose more kinetic energy?case (a) case (b)
Answer:
m1= mass of ist particle = 2g
u1 = initial velocity of ist particle = 8.6 m/s
u2= initial velocity of 2nd particle = 0 ( due to atrest)
m2 = mass of 2nd particle = 1g
In elastic collision ,velocity of ist particle after collisionis given by:
V1 =[ (m1-m2)*u1 + 2*m2*u2] / (m1+m2)
= ( 2*10-3 -1*10-3)*8.6 +2*1*10-3*0 / (2+1)*10-3
= 2.867 m/s
V2 = velocity of 2nd particle after collision
V2 = (2*m1*u1 + ( m2-m1)*u2] / (m1+m2)
= 2*2*10-3*8.6 + ( 1*10-3 -2*10-3)*0 / ( 2+1)*10-3
= 11.47 m/s
——————————————————————
B) now m2 = 10g
V1 = [ (m1-m2)*u1 + 2*m2*u2] / (m1+m2)
= ( 2 – 10)*10-3*8.6 + 2*0 / (2+10)*10-3
= -5.73 m/s
V2 = (2*m1*u1 + ( m2-m1)*u2] / (m1+m2)
= 2*2*10-3*8.6 + ( 10- 2)*10-3*0 / (2+10)*10-3
= 2.867 m/s
—————————————————————————————-
c) final K.E of ist particle in situation a:
K.E = (1/2)*m1*V12
= (1/2)*2*10-3*( 2.867 m/s)2
= 8.219*10-3 J
in situation b:
K.E’ = (1/2)*2*10-3*( -5.73m/s)2
= 0.0328 m/s
in case b particle lose more K.E
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