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A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
- a)mv√2
- b)zero
- c)2mv
- d)mv/√2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A particle of mass m is projected with velocity v making an angle of ...
Given:
- Mass of the particle, m
- Initial velocity of the particle, v
- Angle of projection, θ = 45° with the horizontal
To find:
The magnitude of the change in momentum when the particle lands on the ground.
Explanation:
When a particle is projected at an angle with the horizontal, its horizontal and vertical components of velocity can be determined using trigonometry. In this case, since the angle is 45°, the horizontal and vertical components of velocity will be equal.
Let's calculate the horizontal and vertical components of velocity:
- Horizontal component, Vx = v * cos(θ) = v * cos(45°) = v * (√2/2) = v/√2
- Vertical component, Vy = v * sin(θ) = v * sin(45°) = v * (√2/2) = v/√2
The particle follows a parabolic path due to the effect of gravity. When the particle lands on the ground, its vertical component of velocity becomes zero. However, the horizontal component of velocity remains unchanged.
Since momentum is a vector quantity, the change in momentum of the particle can be calculated by subtracting the initial momentum from the final momentum.
The initial momentum, p_initial = m * v
The final momentum, p_final = m * Vx (since Vy = 0)
Therefore, the change in momentum, Δp = p_final - p_initial = m * Vx - m * v = m * (Vx - v)
Substituting the values of Vx and v:
Δp = m * (v/√2 - v) = m * (v/√2 - √2v/√2) = m * (-√2v/√2) = -m * v
The magnitude of the change in momentum is given by |Δp| = |-m * v| = m * v
Hence, the correct option is (a) mv√2.
- Mass of the particle, m
- Initial velocity of the particle, v
- Angle of projection, θ = 45° with the horizontal
To find:
The magnitude of the change in momentum when the particle lands on the ground.
Explanation:
When a particle is projected at an angle with the horizontal, its horizontal and vertical components of velocity can be determined using trigonometry. In this case, since the angle is 45°, the horizontal and vertical components of velocity will be equal.
Let's calculate the horizontal and vertical components of velocity:
- Horizontal component, Vx = v * cos(θ) = v * cos(45°) = v * (√2/2) = v/√2
- Vertical component, Vy = v * sin(θ) = v * sin(45°) = v * (√2/2) = v/√2
The particle follows a parabolic path due to the effect of gravity. When the particle lands on the ground, its vertical component of velocity becomes zero. However, the horizontal component of velocity remains unchanged.
Since momentum is a vector quantity, the change in momentum of the particle can be calculated by subtracting the initial momentum from the final momentum.
The initial momentum, p_initial = m * v
The final momentum, p_final = m * Vx (since Vy = 0)
Therefore, the change in momentum, Δp = p_final - p_initial = m * Vx - m * v = m * (Vx - v)
Substituting the values of Vx and v:
Δp = m * (v/√2 - v) = m * (v/√2 - √2v/√2) = m * (-√2v/√2) = -m * v
The magnitude of the change in momentum is given by |Δp| = |-m * v| = m * v
Hence, the correct option is (a) mv√2.
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Community Answer
A particle of mass m is projected with velocity v making an angle of ...
The horizontal momentum does not change. The change in vertical momentum is
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A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will bea)mv√2b)zeroc)2mvd)mv/√2Correct answer is option 'A'. Can you explain this answer?
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