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Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be
- a)−0.5ms−1 and 0.3ms−1
- b)0.5ms−1 and −0.3ms−1
- c)−0.3ms−1 and 0.5ms−1
- d)0.3ms−1 and 0.5ms−1
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 ...
Given:
Two identical balls A and B collide elastically in one dimension.
Velocity of ball A before collision = 0.5 m/s
Velocity of ball B before collision = 0.3 m/s
To find:
The velocities of B and A after the collision respectively.
Solution:
When two objects collide elastically, both momentum and kinetic energy are conserved.
Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Let the mass of each ball be 'm'.
Before collision:
Momentum of ball A = mass x velocity = m x 0.5
Momentum of ball B = mass x velocity = m x 0.3
After collision:
Let the velocities of ball A and B after collision be 'v1' and 'v2' respectively.
Momentum of ball A = mass x velocity = m x v1
Momentum of ball B = mass x velocity = m x v2
According to the conservation of momentum:
Momentum before collision = Momentum after collision
m x 0.5 + m x 0.3 = m x v1 + m x v2
0.5 + 0.3 = v1 + v2
0.8 = v1 + v2 ...........(1)
Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Before collision:
Kinetic energy of ball A = (1/2) x mass x velocity^2 = (1/2) x m x (0.5)^2 = 0.125m
Kinetic energy of ball B = (1/2) x mass x velocity^2 = (1/2) x m x (0.3)^2 = 0.045m
After collision:
Kinetic energy of ball A = (1/2) x mass x velocity^2 = (1/2) x m x v1^2 = 0.5m x v1^2
Kinetic energy of ball B = (1/2) x mass x velocity^2 = (1/2) x m x v2^2 = 0.5m x v2^2
According to the conservation of kinetic energy:
Kinetic energy before collision = Kinetic energy after collision
0.125m + 0.045m = 0.5m x v1^2 + 0.5m x v2^2
0.17m = 0.5m x v1^2 + 0.5m x v2^2 ...........(2)
Solving equations (1) and (2):
From equation (1), we have: 0.8 = v1 + v2
Squaring both sides: 0.64 = v1^2 + v2^2 + 2v1v2 ...........(3)
Substituting equation (3) in equation (2):
0.17m = 0.5m x (0.64 - 2v1v2)
0.17 = 0.32 - v1v2
v1v2 = 0.32 - 0.17
v1v2 =
Two identical balls A and B collide elastically in one dimension.
Velocity of ball A before collision = 0.5 m/s
Velocity of ball B before collision = 0.3 m/s
To find:
The velocities of B and A after the collision respectively.
Solution:
When two objects collide elastically, both momentum and kinetic energy are conserved.
Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Let the mass of each ball be 'm'.
Before collision:
Momentum of ball A = mass x velocity = m x 0.5
Momentum of ball B = mass x velocity = m x 0.3
After collision:
Let the velocities of ball A and B after collision be 'v1' and 'v2' respectively.
Momentum of ball A = mass x velocity = m x v1
Momentum of ball B = mass x velocity = m x v2
According to the conservation of momentum:
Momentum before collision = Momentum after collision
m x 0.5 + m x 0.3 = m x v1 + m x v2
0.5 + 0.3 = v1 + v2
0.8 = v1 + v2 ...........(1)
Conservation of kinetic energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Before collision:
Kinetic energy of ball A = (1/2) x mass x velocity^2 = (1/2) x m x (0.5)^2 = 0.125m
Kinetic energy of ball B = (1/2) x mass x velocity^2 = (1/2) x m x (0.3)^2 = 0.045m
After collision:
Kinetic energy of ball A = (1/2) x mass x velocity^2 = (1/2) x m x v1^2 = 0.5m x v1^2
Kinetic energy of ball B = (1/2) x mass x velocity^2 = (1/2) x m x v2^2 = 0.5m x v2^2
According to the conservation of kinetic energy:
Kinetic energy before collision = Kinetic energy after collision
0.125m + 0.045m = 0.5m x v1^2 + 0.5m x v2^2
0.17m = 0.5m x v1^2 + 0.5m x v2^2 ...........(2)
Solving equations (1) and (2):
From equation (1), we have: 0.8 = v1 + v2
Squaring both sides: 0.64 = v1^2 + v2^2 + 2v1v2 ...........(3)
Substituting equation (3) in equation (2):
0.17m = 0.5m x (0.64 - 2v1v2)
0.17 = 0.32 - v1v2
v1v2 = 0.32 - 0.17
v1v2 =
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Community Answer
Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 ...
Masses of the balls are same and collision is elastic, so their velocity will be interchanged after collision.∴ va = +uB = −0.3ms−1 and vB = vA = 0.5ms−1
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Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer?
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Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer?.
Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical balls A and B having velocities of 0.5ms−1 and 0.3ms−1 respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will bea)−0.5ms−1 and 0.3ms−1b)0.5ms−1 and −0.3ms−1c)−0.3ms−1 and 0.5ms−1d)0.3ms−1 and 0.5ms−1Correct answer is option 'B'. Can you explain this answer?.
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