Question

In a pool game, the cue ball, which has an initial speed of 7.0 m/s, make an elastic collision with the eight ball, which is initially at rest. After the collision, the eight ball moves at an angle of 25° to the original direction of the cue ball.

(a) Find the direction of motion of the cue ball after the collision (______) I got the 60° which is right.

(b) Find the speed of each ball. Assume that the balls have equal mass.

(_____m/s)

(_____m/s)

Answer 1

Given that,

Speed = Vicue = 7 m/s ; anlgle = 25 Deg

(a)We need to find the direction of the cue ball fater the collision. Let it be .

We know from conservation of momentum that,

P(i) = P(f) , the intial momentum of the system is due to cue ball only as the other 8 balls are at rest. Let m8 be the combined masses of eight balls and v8 be the velocity. mcue be the mass and Vfcue be the final velocity of cue ball.

mcue x Vicue = mcue x Vfcue + m8 V8

Momentum is conserved in all the directions, Lets assume that the ball is intially moving along Y and the masses are equal, we can write:

- Vfcue cos + V8 cos = 0 (1)

Vfcue sin + V8 sin = Vicue (2)

Squaring 1 and 2 we get;

Vfcue² cos² + V8² cos² - 2 Vfcue V8 cos cos = 0

Vfcue² sin² + V8² sin² + 2 Vfcue V8 sin sin = Vicue²

Adding the above two we get:

Vicue² = Vfcue² + V8² + 2Vfcue V8 cos( + ) (3)

we know that energy is conserved, so

E(i) = E(f)

1/2 mcue Vicue² = 1/2 mcue x Vfcue² + 1/2 m8 V8²

Vicue² =  Vfcue² + V82 (4) (As 1/2 and m gets cancelled both the sides)

Putting the value of 4 in 1 we get:

Vicue² = Vicue² + 2Vfcue V8 cos( + )

( + ) = 90 and we have   = 25 So = 90 - 25 = 65 Deg

Hence, direction of the cue bal after the collision = = 65 Deg.

b)Vfrom the equation we have derived in above parts we have,

V8 = Vicue x cos / sqrt [(cos² + cos²) = 7 x cos(65) / sqrt [(cos25)² + (cos60)²] = 2.86 m/s

Hence, V8 = 2.86 m/s

Vfcue =  Vicue x cos / sqrt [(cos² + cos²) = 7 x cos25 / sqrt [(cos25)² + (cos60)²] = 6.1 m/s

Hence, Vfcue = 6.1 m/s