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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be
- a)11.5 m/s
- b)14.0 m/s
- c)7.0 m/s
- d)9.89 m/s
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A body of mass 5 kg explodes at rest into three fragments with masses ...
If 2 equal masses move in perpendicular direction then 3rd mass will move in direction such at
that it will be resultant of other 2 masses.
hence (3m)^2= (1×21)^2+(1×21)^2
=[(21)^2×(21)^2]^1/2
=[21(2)^1/2 ]\3=7root 2
=9.898
that it will be resultant of other 2 masses.
hence (3m)^2= (1×21)^2+(1×21)^2
=[(21)^2×(21)^2]^1/2
=[21(2)^1/2 ]\3=7root 2
=9.898
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Community Answer
A body of mass 5 kg explodes at rest into three fragments with masses ...
Given:
Mass of body (m) = 5 kg
Fragment masses ratio = 1:1:3
Velocities of equal mass fragments (v) = 21 m/s
To find:
Velocity of the heaviest fragment
Solution:
Let the masses of the three fragments be m1, m2, and m3, respectively.
Given that the masses are in the ratio 1:1:3, we can write:
m1 : m2 : m3 = 1 : 1 : 3
Let the velocities of the fragments be v1, v2, and v3, respectively.
Given that the velocities of the equal mass fragments are 21 m/s, we can write:
v1 = v2 = 21 m/s
Since the body was initially at rest, the total momentum before the explosion is zero.
Therefore, the total momentum after the explosion must also be zero.
Total momentum before the explosion = Total momentum after the explosion
Initial momentum = 0
Final momentum = m1v1 + m2v2 + m3v3
As we know that the masses are in the ratio 1:1:3, we can assume the masses to be x, x, and 3x, respectively.
So, the total momentum after the explosion can be written as:
m1v1 + m2v2 + m3v3 = xv1 + xv2 + 3xv3
= 2(xv1) + 3xv3
= 2(21x) + 3x(21)
= 42x + 63x
= 105x
Since the total momentum before and after the explosion is the same, we have:
105x = 0
x = 0 (Since mass cannot be zero)
This implies that the masses of the fragments are zero, which is not possible.
Therefore, the given scenario is not possible.
Hence, there is no valid answer for the velocity of the heaviest fragment.
Mass of body (m) = 5 kg
Fragment masses ratio = 1:1:3
Velocities of equal mass fragments (v) = 21 m/s
To find:
Velocity of the heaviest fragment
Solution:
Let the masses of the three fragments be m1, m2, and m3, respectively.
Given that the masses are in the ratio 1:1:3, we can write:
m1 : m2 : m3 = 1 : 1 : 3
Let the velocities of the fragments be v1, v2, and v3, respectively.
Given that the velocities of the equal mass fragments are 21 m/s, we can write:
v1 = v2 = 21 m/s
Since the body was initially at rest, the total momentum before the explosion is zero.
Therefore, the total momentum after the explosion must also be zero.
Total momentum before the explosion = Total momentum after the explosion
Initial momentum = 0
Final momentum = m1v1 + m2v2 + m3v3
As we know that the masses are in the ratio 1:1:3, we can assume the masses to be x, x, and 3x, respectively.
So, the total momentum after the explosion can be written as:
m1v1 + m2v2 + m3v3 = xv1 + xv2 + 3xv3
= 2(xv1) + 3xv3
= 2(21x) + 3x(21)
= 42x + 63x
= 105x
Since the total momentum before and after the explosion is the same, we have:
105x = 0
x = 0 (Since mass cannot be zero)
This implies that the masses of the fragments are zero, which is not possible.
Therefore, the given scenario is not possible.
Hence, there is no valid answer for the velocity of the heaviest fragment.
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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will bea)11.5 m/sb)14.0 m/sc)7.0 m/sd)9.89 m/sCorrect answer is option 'D'. Can you explain this answer?
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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will bea)11.5 m/sb)14.0 m/sc)7.0 m/sd)9.89 m/sCorrect answer is option 'D'. 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 of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will bea)11.5 m/sb)14.0 m/sc)7.0 m/sd)9.89 m/sCorrect answer is option 'D'. 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 of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will bea)11.5 m/sb)14.0 m/sc)7.0 m/sd)9.89 m/sCorrect answer is option 'D'. Can you explain this answer?.
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