Question

A 70g ball collides with another ball of 140g the initial velocity of the first ball is 9 m/s to the right and the second ball is at rest if the collision was perfectly elastic what would be the velocities of the balls after collision in m/s ?

Answer 1

Answer:

-3m/s and 6m/s

Explanation:

coefficient of restitution for perfectly elastic collision

,e = 1

formula for coefficient of restitution

V2 -V1= -e(u2-u1) ; V2,V1 velocities after collision

u1,u2 velocities before collision

V2-V1 = -1*(0-9).

V2=9 + V1

applying conservation of momentum

70*u1+140*u2= 70*V1+140V2

70*9 +140*0 = 70*V1+ 140*(9+V1)

630= +70V1 +140*V1 +140*9

210V1= 630-1260

V1= -630/210= -3m/s (towards left)

V2= 9-3=6 m/s

checking the answer with conservation of kinetic energy

.05m1u1^2+.05m2u2^2 == 0.5m1V1^2+0.5m2V2^2

0.5*70*9^2+0.5*140*0^2 =2835g*m^2/s^2 (before collision)

.5*70*(-3)^2+ 0.5*140*6^2 =2835 g*m^2/s^2 (after collision)

Answer 2

Answer:

 .....,

Explanation:

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