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A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' is
- a)1/2
- b)1/3
- c)1/4
- d)1/5
Correct answer is option 'B'. Can you explain this answer?
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Solution:
Given, a perfectly elastic collision between body A and stationary body B.
After the collision, the bodies fly apart in opposite directions with equal velocities.
Let the mass of body A be 'm1' and that of body B be 'm2'.
Let 'u1' and 'u2' be the initial velocities of body A and B, respectively.
After the collision, the velocities become 'v1' and 'v2', respectively.
Since the collision is perfectly elastic, we can use the conservation of momentum and conservation of kinetic energy to solve this problem.
Conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
Since body B is initially at rest, u2 = 0.
Therefore, m1u1 = m1v1 + m2v2
Conservation of kinetic energy:
(1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Since the collision is perfectly elastic, kinetic energy is conserved.
Therefore, (1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Dividing the two equations, we get:
u1^2/v1^2 = (m1 + m2)/m1
Since the bodies fly apart with equal velocities, v1 = v2.
Therefore, u1^2/v1^2 = 2
Substituting v2 = v1 in the conservation of momentum equation, we get:
m1u1 = m1v1 + m2v1
v1 = u1(m1/(m1 + m2))
Substituting v1 = v2 in the equation u1^2/v1^2 = 2, we get:
u1^2/v2^2 = 2
u1/v1 = sqrt(2)
Substituting v1 = u1(m1/(m1 + m2)), we get:
sqrt(2) = (m1 + m2)/m1
m2/m1 = sqrt(2) - 1
m1/m2 = 1/(sqrt(2) - 1)
m1/m2 = 1/((sqrt(2) - 1)(sqrt(2) + 1))
m1/m2 = 1/3
Therefore, the mass ratio of body A and B is 1:3.
Hence, option (b) is the correct answer.
Given, a perfectly elastic collision between body A and stationary body B.
After the collision, the bodies fly apart in opposite directions with equal velocities.
Let the mass of body A be 'm1' and that of body B be 'm2'.
Let 'u1' and 'u2' be the initial velocities of body A and B, respectively.
After the collision, the velocities become 'v1' and 'v2', respectively.
Since the collision is perfectly elastic, we can use the conservation of momentum and conservation of kinetic energy to solve this problem.
Conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
Since body B is initially at rest, u2 = 0.
Therefore, m1u1 = m1v1 + m2v2
Conservation of kinetic energy:
(1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Since the collision is perfectly elastic, kinetic energy is conserved.
Therefore, (1/2)m1u1^2 = (1/2)m1v1^2 + (1/2)m2v2^2
Dividing the two equations, we get:
u1^2/v1^2 = (m1 + m2)/m1
Since the bodies fly apart with equal velocities, v1 = v2.
Therefore, u1^2/v1^2 = 2
Substituting v2 = v1 in the conservation of momentum equation, we get:
m1u1 = m1v1 + m2v1
v1 = u1(m1/(m1 + m2))
Substituting v1 = v2 in the equation u1^2/v1^2 = 2, we get:
u1^2/v2^2 = 2
u1/v1 = sqrt(2)
Substituting v1 = u1(m1/(m1 + m2)), we get:
sqrt(2) = (m1 + m2)/m1
m2/m1 = sqrt(2) - 1
m1/m2 = 1/(sqrt(2) - 1)
m1/m2 = 1/((sqrt(2) - 1)(sqrt(2) + 1))
m1/m2 = 1/3
Therefore, the mass ratio of body A and B is 1:3.
Hence, option (b) is the correct answer.
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A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct answer is option 'B'. Can you explain this answer?
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A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct 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 A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct 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 A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct answer is option 'B'. Can you explain this answer?.
A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct 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 A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct 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 A body 'A' experiences perfectly elastic collision with a stationary body 'B' . If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of 'A' and 'B' isa) 1/2 b) 1/3 c) 1/4 d) 1/5 Correct answer is option 'B'. Can you explain this answer?.
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