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A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is?
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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)
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)
Community Answer
A billiards player hits a stationary ball by an identical ball to pock...
90-35 =55 degree
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A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is?
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A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is?.
A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A billiards player hits a stationary ball by an identical ball to pocket the target ball in A corner pocket that is at an angle of 35 degree with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important,the angle made by the target ball with respect to the incoming ball is?.
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