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A particle of mass 1 kg is projected at an angle 45 degree to the horizontal with an initial velocity of 20m/s. Change in momentum during its flight time is . 1)10√2. 2)20√2. 3)30√2 4)40√2 .?
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A particle of mass 1 kg is projected at an angle 45 degree to the hori...
My answer coming out to be 100√2 not in option but... I used formula L= M ucos@*u^2 sin^2@/2g
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A particle of mass 1 kg is projected at an angle 45 degree to the hori...
Given:
Mass of the particle (m) = 1 kg
Angle of projection (θ) = 45 degrees
Initial velocity (u) = 20 m/s
To Find:
Change in momentum during its flight time
Explanation:
The change in momentum of an object is equal to the product of its mass and the change in velocity.
1. Calculate the horizontal and vertical components of the initial velocity:
The horizontal component (ux) of the initial velocity is given by:
ux = u * cos(θ)
ux = 20 * cos(45)
ux = 20 * (√2 / 2)
ux = 10√2 m/s
The vertical component (uy) of the initial velocity is given by:
uy = u * sin(θ)
uy = 20 * sin(45)
uy = 20 * (√2 / 2)
uy = 10√2 m/s
2. Calculate the time of flight:
The time of flight (T) can be calculated using the vertical component of velocity and acceleration due to gravity:
T = (2 * uy) / g
T = (2 * 10√2) / 9.8
T = (20√2) / 9.8
T ≈ 2.04 s
3. Calculate the change in velocity:
The change in velocity (Δv) can be calculated using the horizontal component of velocity and the time of flight:
Δv = ux * T
Δv = 10√2 * 2.04
Δv ≈ 20√2 m/s
4. Calculate the change in momentum:
The change in momentum (Δp) is equal to the product of the mass and the change in velocity:
Δp = m * Δv
Δp = 1 * (20√2)
Δp ≈ 20√2 kg·m/s
Answer:
Therefore, the change in momentum during the flight time is approximately 20√2 kg·m/s.
Mass of the particle (m) = 1 kg
Angle of projection (θ) = 45 degrees
Initial velocity (u) = 20 m/s
To Find:
Change in momentum during its flight time
Explanation:
The change in momentum of an object is equal to the product of its mass and the change in velocity.
1. Calculate the horizontal and vertical components of the initial velocity:
The horizontal component (ux) of the initial velocity is given by:
ux = u * cos(θ)
ux = 20 * cos(45)
ux = 20 * (√2 / 2)
ux = 10√2 m/s
The vertical component (uy) of the initial velocity is given by:
uy = u * sin(θ)
uy = 20 * sin(45)
uy = 20 * (√2 / 2)
uy = 10√2 m/s
2. Calculate the time of flight:
The time of flight (T) can be calculated using the vertical component of velocity and acceleration due to gravity:
T = (2 * uy) / g
T = (2 * 10√2) / 9.8
T = (20√2) / 9.8
T ≈ 2.04 s
3. Calculate the change in velocity:
The change in velocity (Δv) can be calculated using the horizontal component of velocity and the time of flight:
Δv = ux * T
Δv = 10√2 * 2.04
Δv ≈ 20√2 m/s
4. Calculate the change in momentum:
The change in momentum (Δp) is equal to the product of the mass and the change in velocity:
Δp = m * Δv
Δp = 1 * (20√2)
Δp ≈ 20√2 kg·m/s
Answer:
Therefore, the change in momentum during the flight time is approximately 20√2 kg·m/s.
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A particle of mass 1 kg is projected at an angle 45 degree to the horizontal with an initial velocity of 20m/s. Change in momentum during its flight time is . 1)10√2. 2)20√2. 3)30√2 4)40√2 .?
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A particle of mass 1 kg is projected at an angle 45 degree to the horizontal with an initial velocity of 20m/s. Change in momentum during its flight time is . 1)10√2. 2)20√2. 3)30√2 4)40√2 .? 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 1 kg is projected at an angle 45 degree to the horizontal with an initial velocity of 20m/s. Change in momentum during its flight time is . 1)10√2. 2)20√2. 3)30√2 4)40√2 .? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 1 kg is projected at an angle 45 degree to the horizontal with an initial velocity of 20m/s. Change in momentum during its flight time is . 1)10√2. 2)20√2. 3)30√2 4)40√2 .?.
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