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A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to
- a)
- b)
- c)
- d)3R
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A particle of mass 3m is projected from the ground at some angle with ...
At the time when the particle broke the momentum is conserved as no external force acted
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
Most Upvoted Answer
A particle of mass 3m is projected from the ground at some angle with ...
Let the angle of projection be A and velocity of projection be "u".
Range, R = (2u²sin A cos A)/g
At the highest point only horizontal component of velocity is present. On applying conservation of momentum at this point,
(3m)(ucos A) = (2m)v + m(0)
=> v = 3ucos A/2
The height of particle of mass 2m is,
H = u²sin²A/2g
From here let the particle take time "t" to come down,
H = gt²/2
u²sin²A/2g = gt²/2
t = usin A/g
From the highest point the horizontal displacement of the particle is d = vt
d = (usin A/g)(3ucos A/2)
d = (3u²sin A cos A)/2g
=> d = 3R/4
At the time of breaking the particles are at a distance R/2 from the point of projection. So, the final distance of the particle from the point of projection,
x = R/2 + d
=> x = R/2 + 3R/4
=> x = 5R/4
Range, R = (2u²sin A cos A)/g
At the highest point only horizontal component of velocity is present. On applying conservation of momentum at this point,
(3m)(ucos A) = (2m)v + m(0)
=> v = 3ucos A/2
The height of particle of mass 2m is,
H = u²sin²A/2g
From here let the particle take time "t" to come down,
H = gt²/2
u²sin²A/2g = gt²/2
t = usin A/g
From the highest point the horizontal displacement of the particle is d = vt
d = (usin A/g)(3ucos A/2)
d = (3u²sin A cos A)/2g
=> d = 3R/4
At the time of breaking the particles are at a distance R/2 from the point of projection. So, the final distance of the particle from the point of projection,
x = R/2 + d
=> x = R/2 + 3R/4
=> x = 5R/4
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Community Answer
A particle of mass 3m is projected from the ground at some angle with ...
At the time when the particle broke the momentum is conserved as no external force acted
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
As the vertical component was zero and let say horizontal component was v
Thus we get,
3mv = m.0 + 2m.u
Thus we get u = 3/2 . v
But as there was no change in vertical component of velocity, time of flight does not change. Thus the distance at which the particle would land would be t x v + t x u
Where 2t is time of flight and 2t x v = R
Thus x = R/2 + 3R/4
= 5R/4
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A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer?
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A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer?.
A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer?.
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A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A particle of mass 3m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal toa)b)c)d)3RCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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