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A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite point
  • a)
    The momentum changes by mυ
  • b)
    The momentum changes by 2mυ
  • c)
    The kinetic energy changes by (1/2)mυ2
  • d)
    The kinetic energy changes by mυ2
Correct answer is option 'B'. Can you explain this answer?
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A particle is moving in a circle with uniform speed υ. In movi...
Momentum changes by 2mv but kinetic energy remains same.
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A particle is moving in a circle with uniform speed υ. In movi...
If a particle is moving in a circle with uniform speed, it means that the magnitude of its velocity is constant throughout the motion. However, the direction of its velocity is continuously changing as it moves around the circle.

This type of motion is called uniform circular motion. In uniform circular motion, the particle's acceleration is directed towards the center of the circle and is always perpendicular to the velocity vector. The magnitude of the acceleration is given by the equation:

a = (v^2) / r

where a is the magnitude of the acceleration, v is the magnitude of the velocity, and r is the radius of the circle.

Since the speed is constant, the magnitude of the velocity is also constant, so the acceleration is solely determined by the radius of the circle. The smaller the radius, the larger the acceleration required to keep the particle moving in a circle.

It is important to note that even though the speed is constant, the particle is still accelerating because acceleration is defined as any change in velocity, including changes in direction.
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A particle is moving in a circle with uniform speed υ. In movi...

Initial velocity v1 = v
Final velocity v2 = -v
Initial momentum p1 = mv
Final momentum p2 = m(-v) = -mv
Change in momentum
Δp = p1 - p2
⇒ mv - (-mv)
⇒ 2mv
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Can you explain the answer of this question below:The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q.The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is :A:H D TB:H = D = TC:H D TD:H D TThe answer is a.

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A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite pointa)The momentum changes by mυb)The momentum changes by 2mυc)The kinetic energy changes by (1/2)mυ2d)The kinetic energy changes by mυ2Correct answer is option 'B'. Can you explain this answer?
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A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite pointa)The momentum changes by mυb)The momentum changes by 2mυc)The kinetic energy changes by (1/2)mυ2d)The kinetic energy changes by mυ2Correct answer is option 'B'. 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 A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite pointa)The momentum changes by mυb)The momentum changes by 2mυc)The kinetic energy changes by (1/2)mυ2d)The kinetic energy changes by mυ2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle is moving in a circle with uniform speed υ. In moving from a point to another diametrically opposite pointa)The momentum changes by mυb)The momentum changes by 2mυc)The kinetic energy changes by (1/2)mυ2d)The kinetic energy changes by mυ2Correct answer is option 'B'. Can you explain this answer?.
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