A billiards player hits a stationary ball by an identical ball to pock...
Explanation:
Given:
Angle of corner pocket = 35°
Assuming the collision as elastic
Friction and rotational motion are not important
To Find:
Angle made by the target ball with respect to the incoming ball
Solution:
Let us assume that the first ball is moving with a velocity V before the collision.
After the collision, the first ball will move in a direction making an angle 35° with the initial direction.
Let us assume that the second ball is released from rest.
Now, using the principle of conservation of momentum, we can write:
Momentum before collision = Momentum after collision
m1V = m1V1cosθ + m2V2cosϕ
Here, m1 and m2 are the masses of the two balls, V1 is the velocity of the first ball after the collision, V2 is the velocity of the second ball after the collision, θ is the angle made by the first ball after the collision with the initial direction, and ϕ is the angle made by the second ball after the collision with the initial direction.
Since the second ball is released from rest, V2 = V1.
Also, θ = 35°.
Solving the equation, we get:
cosϕ = (m1V - m1V1cosθ) / m2V1
cosϕ = (m1 - m1cosθ) / m2
cosϕ = m1(1 - cosθ) / m2
Now, using the principle of conservation of energy, we can write:
Kinetic energy before collision = Kinetic energy after collision
(1/2)m1V^2 = (1/2)m1V1^2 + (1/2)m2V2^2
Since the collision is elastic, we can write:
m1V^2 = m1V1^2 + m2V2^2
Substituting V2 = V1, we get:
m1V^2 = m1V1^2 + m2V1^2
V1^2 = (m1V^2) / (m1 + m2)
Substituting this value in the expression for cosϕ, we get:
cosϕ = m1(1 - cosθ) / m2
= m1(1 - cos35°) / m2
= 0.305m1 / m2
Taking the inverse cosine, we get:
ϕ = cos^-1(0.305m1 / m2)
Hence, the angle made by the target ball with respect to the incoming ball is given by:
ϕ = cos^-1(0.305m1 / m2)
A billiards player hits a stationary ball by an identical ball to pock...
90-35 =55 degree
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