An object is projected with a speed 10 m s - 1 at an angle of 30 ° with the horizontal. It breaks into n equal fragments during its motion. One fragment strikes the ground at a distance of 3 m from the point of projection. The centre of mass of the remaining fragments strikes the ground at a distance of 7 3 m from the point of projection. If all fragments strike the ground at the same time, Find the value of n .

Position of overall center of mass of the system, R c m = 1 0 2 sin ⁡ 6 0 g = 5 3 m If m is the mass of one fragment then, 5 3 = m 3 + m ( n - 1 ) 7 3 m n n = 3

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Physics - Rotation Question with Solution | TestHub

PhysicsRotationMiscellaneousEasy2 minPYQ_2020

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An object is projected with a speed $10 m s^{- 1}$ at an angle of $30 °$ with the horizontal. It breaks into $n$ equal fragments during its motion. One fragment strikes the ground at a distance of $\sqrt{3} m$ from the point of projection. The centre of mass of the remaining fragments strikes the ground at a distance of $7 \sqrt{3} m$ from the point of projection. If all fragments strike the ground at the same time, Find the value of $n$ .

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Answer:

Solution:

Position of overall center of mass of the system,

$R_{c m} = \frac{1 0^{2} \sin 6 0}{g} = 5 \sqrt{3} m$

If $m$ is the mass of one fragment then,

$5 \sqrt{3} = \frac{m \sqrt{3} + m (n - 1) 7 \sqrt{3}}{m n}$

$n = 3$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:RotationSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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