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A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent?
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A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass ...
Given:
Mass of the sand bag (m1) = 1.9 kg
Mass of the bullet (m2) = 100 g = 0.1 kg
Vertical height (h) = 250 m
To find:
Initial velocity of the bullet (u)
Assumptions:
1. There is no air resistance.
2. The collision between the bullet and the sand bag is elastic, i.e., no energy is lost in the collision.
Analysis:
1. Conservation of momentum:
Before the collision, the bullet is moving with velocity u and the sand bag is at rest. After the collision, the bullet gets embedded into the sand bag and they move together with a common velocity v.
According to the law of conservation of momentum,
m1 * 0 + m2 * u = (m1 + m2) * v
2. Conservation of energy:
The initial total energy is equal to the final total energy. Initially, only the bullet has kinetic energy, and finally, the bullet-sand bag system has both kinetic and potential energy.
Initial kinetic energy of the bullet = (1/2) * m2 * u^2
Final kinetic energy of the bullet-sand bag system = (1/2) * (m1 + m2) * v^2
Final potential energy of the bullet-sand bag system = (m1 + m2) * g * h
3. Solving the equations:
Equating the initial and final kinetic energies,
(1/2) * m2 * u^2 = (1/2) * (m1 + m2) * v^2
Equating the final kinetic and potential energies,
(1/2) * (m1 + m2) * v^2 + (m1 + m2) * g * h = (1/2) * m2 * u^2
Simplifying the equations,
m2 * u^2 = (m1 + m2) * v^2
(m1 + m2) * v^2 + (m1 + m2) * g * h = m2 * u^2
Dividing the two equations,
v^2 + g * h = u^2
Substituting the given values,
(0 + 9.8 * 250) = u^2
u^2 = 2450
u ≈ √2450
u ≈ 49.5 m/s
Answer:
The initial velocity of the bullet is approximately 49.5 m/s.
Mass of the sand bag (m1) = 1.9 kg
Mass of the bullet (m2) = 100 g = 0.1 kg
Vertical height (h) = 250 m
To find:
Initial velocity of the bullet (u)
Assumptions:
1. There is no air resistance.
2. The collision between the bullet and the sand bag is elastic, i.e., no energy is lost in the collision.
Analysis:
1. Conservation of momentum:
Before the collision, the bullet is moving with velocity u and the sand bag is at rest. After the collision, the bullet gets embedded into the sand bag and they move together with a common velocity v.
According to the law of conservation of momentum,
m1 * 0 + m2 * u = (m1 + m2) * v
2. Conservation of energy:
The initial total energy is equal to the final total energy. Initially, only the bullet has kinetic energy, and finally, the bullet-sand bag system has both kinetic and potential energy.
Initial kinetic energy of the bullet = (1/2) * m2 * u^2
Final kinetic energy of the bullet-sand bag system = (1/2) * (m1 + m2) * v^2
Final potential energy of the bullet-sand bag system = (m1 + m2) * g * h
3. Solving the equations:
Equating the initial and final kinetic energies,
(1/2) * m2 * u^2 = (1/2) * (m1 + m2) * v^2
Equating the final kinetic and potential energies,
(1/2) * (m1 + m2) * v^2 + (m1 + m2) * g * h = (1/2) * m2 * u^2
Simplifying the equations,
m2 * u^2 = (m1 + m2) * v^2
(m1 + m2) * v^2 + (m1 + m2) * g * h = m2 * u^2
Dividing the two equations,
v^2 + g * h = u^2
Substituting the given values,
(0 + 9.8 * 250) = u^2
u^2 = 2450
u ≈ √2450
u ≈ 49.5 m/s
Answer:
The initial velocity of the bullet is approximately 49.5 m/s.
Community Answer
A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass ...
1400 m/s
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A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent?
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A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent?.
A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bag of sand of mass 1.9 kg is suspended by a rope. A bullet of mass 100g is fired at it and gets embedded into it. The bag rises up a vertical height of 250 m. The initial velocity of the bullet is nearly- (a)700m/s (b)1400m/s (c)2100m/s (d)2800m/s pls answer this it's urgent?.
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