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A body of mass 5 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass 'm' are found to move with a speed of 60 m/s is mutually perpendicular directions. The velocity of third fragment is
- a)60 √2
- b)20 √3
- c)10 √2
- d)20 √2
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A body of mass 5 m initially at rest explodes into 3 fragments with ma...
Mass of two fragments are m. Therefore mass of third fragment is equal to 3xm=3m
Let us assume the direction such that one particle of mass m is moving with 60m/s in +x-direction and other in +y-direction with 60m/s.
Let velocity of third particle be v.
Applying the principle of conservation of linear momentum:
Most Upvoted Answer
A body of mass 5 m initially at rest explodes into 3 fragments with ma...
To solve this problem, we can use the principle of conservation of momentum.
Let the initial mass of the body be M and the mass of each fragment be m. Since the mass ratio of the fragments is 3:1:1, we have:
m1 = 3m
m2 = m
m3 = m
Let v1 be the velocity of fragment 1, v2 be the velocity of fragment 2, and v3 be the velocity of fragment 3.
According to the principle of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion.
Initially, the body is at rest, so the total momentum before the explosion is 0.
After the explosion, the total momentum can be calculated as follows:
Total momentum after explosion = m1 * v1 + m2 * v2 + m3 * v3
Since fragments 2 and 3 each have a mass of m and move with a speed of 60 m/s in mutually perpendicular directions, the total momentum after the explosion can be written as:
Total momentum after explosion = m1 * v1 + (m2 * (-60 m/s)) + (m3 * 60 m/s)
Since the total momentum before and after the explosion should be equal, we have:
0 = m1 * v1 + (m2 * (-60 m/s)) + (m3 * 60 m/s)
Substituting the values of m1, m2, and m3, we have:
0 = (3m) * v1 + (m * (-60 m/s)) + (m * 60 m/s)
Simplifying the equation, we get:
0 = 3mv1 - 60m + 60m
0 = 3mv1
Since m is not equal to 0, we can divide both sides of the equation by 3m:
0 = v1
Therefore, the velocity of the third fragment (v3) is 0 m/s.
Let the initial mass of the body be M and the mass of each fragment be m. Since the mass ratio of the fragments is 3:1:1, we have:
m1 = 3m
m2 = m
m3 = m
Let v1 be the velocity of fragment 1, v2 be the velocity of fragment 2, and v3 be the velocity of fragment 3.
According to the principle of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion.
Initially, the body is at rest, so the total momentum before the explosion is 0.
After the explosion, the total momentum can be calculated as follows:
Total momentum after explosion = m1 * v1 + m2 * v2 + m3 * v3
Since fragments 2 and 3 each have a mass of m and move with a speed of 60 m/s in mutually perpendicular directions, the total momentum after the explosion can be written as:
Total momentum after explosion = m1 * v1 + (m2 * (-60 m/s)) + (m3 * 60 m/s)
Since the total momentum before and after the explosion should be equal, we have:
0 = m1 * v1 + (m2 * (-60 m/s)) + (m3 * 60 m/s)
Substituting the values of m1, m2, and m3, we have:
0 = (3m) * v1 + (m * (-60 m/s)) + (m * 60 m/s)
Simplifying the equation, we get:
0 = 3mv1 - 60m + 60m
0 = 3mv1
Since m is not equal to 0, we can divide both sides of the equation by 3m:
0 = v1
Therefore, the velocity of the third fragment (v3) is 0 m/s.
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Community Answer
A body of mass 5 m initially at rest explodes into 3 fragments with ma...
Apply law of conservation of momentum
initially momentum is zero and afterwards momentum will be -
by vector addition of two momentum vectors mutually perpendicular we get net momentum of 60√2m.and the 3rd particle of mass 3m will have momentum opposite to net momentum of two mutually perpendicular momentum vectors. by equating 60√2m=3mV we get velocity of 3m mass - 20√2
initially momentum is zero and afterwards momentum will be -
by vector addition of two momentum vectors mutually perpendicular we get net momentum of 60√2m.and the 3rd particle of mass 3m will have momentum opposite to net momentum of two mutually perpendicular momentum vectors. by equating 60√2m=3mV we get velocity of 3m mass - 20√2
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A body of mass 5 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass m are found to move with a speed of 60 m/s is mutually perpendicular directions. The velocity of third fragment isa)60 √2b)20 √3c)10 √2d)20 √2Correct answer is option 'D'. Can you explain this answer?
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A body of mass 5 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass m are found to move with a speed of 60 m/s is mutually perpendicular directions. The velocity of third fragment isa)60 √2b)20 √3c)10 √2d)20 √2Correct 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 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass m are found to move with a speed of 60 m/s is mutually perpendicular directions. The velocity of third fragment isa)60 √2b)20 √3c)10 √2d)20 √2Correct 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 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. Two of fragments each of mass m are found to move with a speed of 60 m/s is mutually perpendicular directions. The velocity of third fragment isa)60 √2b)20 √3c)10 √2d)20 √2Correct answer is option 'D'. Can you explain this answer?.
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