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Two particles A and B start moving due to their mutual interaction only. If at any timet, aA and aB are their respective accelerations, vA and vB are their respective velocities, and up to that time WA and WB are the work done on A and B respectively by the mutual force, mA and mB are their masses respectively, then which of the following is always correct?
- a)mAvA +mBvB=0
- b)vA+vB=0
- c)WA+WB=0
- d)aA+aB=0
Correct answer is option 'A'. Can you explain this answer?
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Two particlesAandBstart moving due to their mutual interaction only. I...
Given information:
- Two particles A and B are moving due to their mutual interaction only.
- At any time t, aA and aB are their respective accelerations, vA and vB are their respective velocities, WA and WB are the work done on A and B respectively by the mutual force, mA and mB are their masses respectively.
To find:
Which of the following is always correct?
Solution:
We know that the work done by a force on a particle is given by the product of the force and the displacement of the particle in the direction of the force. Mathematically,
Work = Force x Displacement x cosθ
where θ is the angle between the force and the displacement vectors.
Now, let us consider the two particles A and B.
Work done by the mutual force on particle A:
- The force on particle A is due to particle B and vice versa.
- Let F be the force exerted by particle B on A. Then, the force exerted by A on B will be -F (action-reaction pair).
- Let dA be the displacement of particle A due to this force. Then, the displacement of particle B will be -dB (opposite direction).
- The work done by B on A will be WA = F x dA x cosθ1, where θ1 is the angle between F and dA.
- The work done by A on B will be -WB = -(-F) x (-dB) x cosθ2 = F x dB x cosθ2, where θ2 is the angle between -F and -dB (or F and dB).
Multiplying the above two equations, we get:
WA x (-WB) = F x dA x cosθ1 x F x dB x cosθ2
Simplifying, we get:
WA x WB = F^2 x dA x dB x cosθ1 x cosθ2
Now, let us consider the accelerations of the particles.
Newton's second law states that the net force on a particle is equal to its mass times acceleration. Mathematically,
F = m x a
where F is the net force, m is the mass of the particle, and a is its acceleration.
For particle A, we have:
F = mB x aA
For particle B, we have:
F = mA x aB
Substituting these values in the earlier equation, we get:
WA x WB = (mB x aA)^2 x dA x dB x cosθ1 x cosθ2
Dividing both sides by mAmB, we get:
(WA/mA) x (WB/mB) = aA^2 x dA x dB x cosθ1 x cosθ2
Taking square root on both sides, we get:
WA/mA x WB/mB = ±aA x vA x aB x vB
where ± indicates that the product can be positive or negative depending on the signs of aA, vA, aB, and vB.
Now, let us analyze the options given:
a) mAvA mBvB = 0
Multiplying the two velocities, we get the product of the magnitudes and the cosine of the angle between them. This product can be zero if either of the velocities is zero or if the angle between them is 90 degrees. However, this does
- Two particles A and B are moving due to their mutual interaction only.
- At any time t, aA and aB are their respective accelerations, vA and vB are their respective velocities, WA and WB are the work done on A and B respectively by the mutual force, mA and mB are their masses respectively.
To find:
Which of the following is always correct?
Solution:
We know that the work done by a force on a particle is given by the product of the force and the displacement of the particle in the direction of the force. Mathematically,
Work = Force x Displacement x cosθ
where θ is the angle between the force and the displacement vectors.
Now, let us consider the two particles A and B.
Work done by the mutual force on particle A:
- The force on particle A is due to particle B and vice versa.
- Let F be the force exerted by particle B on A. Then, the force exerted by A on B will be -F (action-reaction pair).
- Let dA be the displacement of particle A due to this force. Then, the displacement of particle B will be -dB (opposite direction).
- The work done by B on A will be WA = F x dA x cosθ1, where θ1 is the angle between F and dA.
- The work done by A on B will be -WB = -(-F) x (-dB) x cosθ2 = F x dB x cosθ2, where θ2 is the angle between -F and -dB (or F and dB).
Multiplying the above two equations, we get:
WA x (-WB) = F x dA x cosθ1 x F x dB x cosθ2
Simplifying, we get:
WA x WB = F^2 x dA x dB x cosθ1 x cosθ2
Now, let us consider the accelerations of the particles.
Newton's second law states that the net force on a particle is equal to its mass times acceleration. Mathematically,
F = m x a
where F is the net force, m is the mass of the particle, and a is its acceleration.
For particle A, we have:
F = mB x aA
For particle B, we have:
F = mA x aB
Substituting these values in the earlier equation, we get:
WA x WB = (mB x aA)^2 x dA x dB x cosθ1 x cosθ2
Dividing both sides by mAmB, we get:
(WA/mA) x (WB/mB) = aA^2 x dA x dB x cosθ1 x cosθ2
Taking square root on both sides, we get:
WA/mA x WB/mB = ±aA x vA x aB x vB
where ± indicates that the product can be positive or negative depending on the signs of aA, vA, aB, and vB.
Now, let us analyze the options given:
a) mAvA mBvB = 0
Multiplying the two velocities, we get the product of the magnitudes and the cosine of the angle between them. This product can be zero if either of the velocities is zero or if the angle between them is 90 degrees. However, this does
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Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer?
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Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer?.
Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer?.
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Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Two particlesAandBstart moving due to their mutual interaction only. If at any timet,aAand aBare their respective accelerations,vAand vBare their respective velocities, and up to that timeWAand WBare the work done onAandBrespectively by the mutual force,mAand mBare their masses respectively, then which of the following is always correct?a)mAvA+mBvB=0b)vA+vB=0c)WA+WB=0d)aA+aB=0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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