One billiard ball is shot east at 2.2 m/s. a second, identical billiard ball is shot west at 0.80 m/s. the balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 degrees and sending it north at 1.36 m/s.

a. what is the speed of the first ball after the collision?

b. what is the direction of the first ball after the collision? give the direction as an angle south of east.

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

Along X- axis

Put the value into the formula

Along Y-axis

Put the value into the formula

Then the final speed of the first ball

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j