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Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)
Correct answer is '2.2'. Can you explain this answer?
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Verified Answer
Two balls, of mass m and mass 2m. approach from perpendicular directio...
Applying the law of conservation
of momentum along x-axis,
Applying the law of conservation of momentum along y-axis
Dividing (ii) by (i).
⇒ α = 2.2
of momentum along x-axis,
Applying the law of conservation of momentum along y-axis
Dividing (ii) by (i).
⇒ α = 2.2
Most Upvoted Answer
Two balls, of mass m and mass 2m. approach from perpendicular directio...
To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy.
Let's assume the initial speed of both balls is denoted as V.
1. Conservation of Momentum:
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Initial momentum:
The momentum of the first ball (mass m) before the collision is given by:
Momentum_1 = mass_1 * velocity_1 = m * V
The momentum of the second ball (mass 2m) before the collision is given by:
Momentum_2 = mass_2 * velocity_2 = (2m) * V
Total initial momentum = Momentum_1 + Momentum_2 = mV + 2mV = 3mV
After the collision:
The more massive ball (mass 2m) moves downward with speed v.
Momentum_1_prime = mass_1 * velocity_1_prime = (2m) * v (downward direction)
The less massive ball (mass m) moves with speed U at an angle θ.
Momentum_2_prime = mass_2 * velocity_2_prime = m * U * cos(θ) (horizontal direction)
Total final momentum = Momentum_1_prime + Momentum_2_prime = 2mv (downward) + mUcos(θ) (horizontal)
According to the conservation of momentum:
Total initial momentum = Total final momentum
3mV = 2mv + mUcos(θ) ...(Equation 1)
2. Conservation of Kinetic Energy:
According to the conservation of kinetic energy, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.
Initial kinetic energy:
The kinetic energy of the first ball (mass m) before the collision is given by:
Kinetic Energy_1 = (1/2) * mass_1 * velocity_1^2 = (1/2) * m * V^2
The kinetic energy of the second ball (mass 2m) before the collision is given by:
Kinetic Energy_2 = (1/2) * mass_2 * velocity_2^2 = (1/2) * (2m) * V^2 = m * V^2
Total initial kinetic energy = Kinetic Energy_1 + Kinetic Energy_2 = (1/2) * m * V^2 + m * V^2 = (3/2) * m * V^2
After the collision:
The more massive ball (mass 2m) moves downward with speed v.
Kinetic Energy_1_prime = (1/2) * mass_1 * velocity_1_prime^2 = (1/2) * (2m) * v^2 = m * v^2
The less massive ball (mass m) moves with speed U at an angle θ.
Kinetic Energy_2_prime = (1/2) * mass_2 * velocity_2_prime^2 = (1/2) * m * U^2
Total final kinetic energy = Kinetic Energy_1_prime + Kinetic Energy_2_prime = m * v^2 + (1/2) * m * U^2
According to the conservation of kinetic energy:
Total initial kinetic energy = Total final kinetic energy
(3
Let's assume the initial speed of both balls is denoted as V.
1. Conservation of Momentum:
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Initial momentum:
The momentum of the first ball (mass m) before the collision is given by:
Momentum_1 = mass_1 * velocity_1 = m * V
The momentum of the second ball (mass 2m) before the collision is given by:
Momentum_2 = mass_2 * velocity_2 = (2m) * V
Total initial momentum = Momentum_1 + Momentum_2 = mV + 2mV = 3mV
After the collision:
The more massive ball (mass 2m) moves downward with speed v.
Momentum_1_prime = mass_1 * velocity_1_prime = (2m) * v (downward direction)
The less massive ball (mass m) moves with speed U at an angle θ.
Momentum_2_prime = mass_2 * velocity_2_prime = m * U * cos(θ) (horizontal direction)
Total final momentum = Momentum_1_prime + Momentum_2_prime = 2mv (downward) + mUcos(θ) (horizontal)
According to the conservation of momentum:
Total initial momentum = Total final momentum
3mV = 2mv + mUcos(θ) ...(Equation 1)
2. Conservation of Kinetic Energy:
According to the conservation of kinetic energy, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.
Initial kinetic energy:
The kinetic energy of the first ball (mass m) before the collision is given by:
Kinetic Energy_1 = (1/2) * mass_1 * velocity_1^2 = (1/2) * m * V^2
The kinetic energy of the second ball (mass 2m) before the collision is given by:
Kinetic Energy_2 = (1/2) * mass_2 * velocity_2^2 = (1/2) * (2m) * V^2 = m * V^2
Total initial kinetic energy = Kinetic Energy_1 + Kinetic Energy_2 = (1/2) * m * V^2 + m * V^2 = (3/2) * m * V^2
After the collision:
The more massive ball (mass 2m) moves downward with speed v.
Kinetic Energy_1_prime = (1/2) * mass_1 * velocity_1_prime^2 = (1/2) * (2m) * v^2 = m * v^2
The less massive ball (mass m) moves with speed U at an angle θ.
Kinetic Energy_2_prime = (1/2) * mass_2 * velocity_2_prime^2 = (1/2) * m * U^2
Total final kinetic energy = Kinetic Energy_1_prime + Kinetic Energy_2_prime = m * v^2 + (1/2) * m * U^2
According to the conservation of kinetic energy:
Total initial kinetic energy = Total final kinetic energy
(3
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Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer?
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Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer?.
Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two balls, of mass m and mass 2m. approach from perpendicular directions with identical speeds and collide. After the collision, the more massive ball moves with the same speed v but downward, perpendicular to its original direction. The less massive ball moves with speed U at an angle θ with respect to the horizontal. If no external forces act during the collision then the final speed of less massive ball is αv. The value of α is_______ (up to 1 decimal place)Correct answer is '2.2'. Can you explain this answer?.
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