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A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :
- a)The position coordinate of particle are (1/2 , 1/4)
- b)The velocity of particle will be along the line 4x − 4y − 1 = 0
- c)The magnitude of velocity at that instant is 4√2 m s−1
- d)The magnitude of angular momentum of particle about origin at that position is 0.
Correct answer is option 'A,B,C'. Can you explain this answer?
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Most Upvoted Answer
A particle of mass m moves along a curve y = x2. When particle has x-...
On the curve
y = x2 ⇒ at x = 1/2
y = 1/4
Hence the coordinate 
Differentiating : y = x2 ⇒ vy = 2xvx
Which satisfies the line
4x − 4y − 1= 0 (tangent to the curve)
and magnitude of velocity :
As the line 4x - 4y = 1 does not pass through the origin, therefore The magnitude of angular momentum of particle about origin at that position is 0 is not correct.
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A particle of mass m moves along a curve y = x2. When particle has x-...
Given:
- Curve: y = x^2
- x-coordinate = 1/2
- x-component of velocity = 4 m/s
To Find:
a) The position coordinates of the particle
b) The velocity of the particle
c) The magnitude of velocity at that instant
d) The magnitude of angular momentum of the particle about the origin at that position
Solution:
a) Position Coordinates of the Particle:
Given that the particle moves along the curve y = x^2, we can substitute the x-coordinate (1/2) into the equation to find the y-coordinate.
Substituting x = 1/2 into y = x^2:
y = (1/2)^2
y = 1/4
Therefore, the position coordinates of the particle are (1/2, 1/4).
b) Velocity of the Particle:
We are given that the x-component of velocity is 4 m/s. The velocity of the particle can be represented as a vector with components (v_x, v_y), where v_x is the x-component and v_y is the y-component of velocity.
Since the particle moves along the curve y = x^2, the y-component of velocity can be found by taking the derivative of y with respect to x.
Differentiating y = x^2 with respect to x:
dy/dx = 2x
The y-component of velocity (v_y) is given by:
v_y = dy/dx * v_x
Substituting the values:
v_y = 2x * v_x
v_y = 2(1/2) * 4
v_y = 4 m/s
Therefore, the velocity of the particle is (4 m/s, 4 m/s).
c) Magnitude of Velocity:
The magnitude of velocity can be found using the formula:
|v| = sqrt(v_x^2 + v_y^2)
Substituting the values:
|v| = sqrt((4)^2 + (4)^2)
|v| = sqrt(16 + 16)
|v| = sqrt(32)
|v| = 4√2 m/s
Therefore, the magnitude of velocity at that instant is 4√2 m/s.
d) Magnitude of Angular Momentum:
The magnitude of angular momentum of a particle about the origin is given by the formula:
L = m * r * v
Since the particle is moving along the curve y = x^2, the radial distance (r) can be found by taking the square root of the sum of squares of x and y.
r = sqrt(x^2 + y^2)
r = sqrt((1/2)^2 + (1/4)^2)
r = sqrt(1/4 + 1/16)
r = sqrt(5/16)
r = sqrt(5)/4
Substituting the values:
L = m * r * |v|
L = m * (sqrt(5)/4) * (4√2)
L = √5 * m
Since the magnitude of angular momentum is given as 0, we have:
√5 * m = 0
This implies that m = 0.
Therefore, the magnitude of angular momentum of the particle
- Curve: y = x^2
- x-coordinate = 1/2
- x-component of velocity = 4 m/s
To Find:
a) The position coordinates of the particle
b) The velocity of the particle
c) The magnitude of velocity at that instant
d) The magnitude of angular momentum of the particle about the origin at that position
Solution:
a) Position Coordinates of the Particle:
Given that the particle moves along the curve y = x^2, we can substitute the x-coordinate (1/2) into the equation to find the y-coordinate.
Substituting x = 1/2 into y = x^2:
y = (1/2)^2
y = 1/4
Therefore, the position coordinates of the particle are (1/2, 1/4).
b) Velocity of the Particle:
We are given that the x-component of velocity is 4 m/s. The velocity of the particle can be represented as a vector with components (v_x, v_y), where v_x is the x-component and v_y is the y-component of velocity.
Since the particle moves along the curve y = x^2, the y-component of velocity can be found by taking the derivative of y with respect to x.
Differentiating y = x^2 with respect to x:
dy/dx = 2x
The y-component of velocity (v_y) is given by:
v_y = dy/dx * v_x
Substituting the values:
v_y = 2x * v_x
v_y = 2(1/2) * 4
v_y = 4 m/s
Therefore, the velocity of the particle is (4 m/s, 4 m/s).
c) Magnitude of Velocity:
The magnitude of velocity can be found using the formula:
|v| = sqrt(v_x^2 + v_y^2)
Substituting the values:
|v| = sqrt((4)^2 + (4)^2)
|v| = sqrt(16 + 16)
|v| = sqrt(32)
|v| = 4√2 m/s
Therefore, the magnitude of velocity at that instant is 4√2 m/s.
d) Magnitude of Angular Momentum:
The magnitude of angular momentum of a particle about the origin is given by the formula:
L = m * r * v
Since the particle is moving along the curve y = x^2, the radial distance (r) can be found by taking the square root of the sum of squares of x and y.
r = sqrt(x^2 + y^2)
r = sqrt((1/2)^2 + (1/4)^2)
r = sqrt(1/4 + 1/16)
r = sqrt(5/16)
r = sqrt(5)/4
Substituting the values:
L = m * r * |v|
L = m * (sqrt(5)/4) * (4√2)
L = √5 * m
Since the magnitude of angular momentum is given as 0, we have:
√5 * m = 0
This implies that m = 0.
Therefore, the magnitude of angular momentum of the particle
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A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer?
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A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer?.
A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer?.
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ample number of questions to practice A particle of mass m moves along a curve y = x2. When particle has x-coordinate as 1/2 and x−component of velocity as 4 m s−1 then :a)The position coordinate of particle are (1/2 , 1/4)b)The velocity of particle will be along the line 4x − 4y − 1 = 0c)The magnitude of velocity at that instant is 4√2 m s−1d)The magnitude of angular momentum of particle about origin at that position is 0.Correct answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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