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Standard XIII

Physics

Projectile Time, Height and Range

A particle of...

Question

A particle of mass $1 k g$ is projected with a velocity of $20 m / s$ at an angle of $30^{\circ}$ with the horizontal. Find the change in the momentum $(k g - m / s)$ of the particle between the points of projection and the highest point. Take $g = 10 m / s^{2}$

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Solution

The correct option is B $10$
Since the velocity is constant in the horizontal direction, momentum will change only in vertical direction. At the highest point, velocity in the vertical direction is zero.
Hence, $Δ p = m u sin θ = 1 \times 20 \times \frac{1}{2} = 10 k g - m / s$

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A particle of mass 1 kg is projected with a velocity of 20 m/s at an angle of 30∘ with the horizontal. Find the change in the momentum (kg−m/s) of the particle between the points of projection and the highest point. Take g=10 m/s2

The correct option is B 10Since the velocity is constant in the horizontal direction, momentum will change only in vertical direction. At the highest point, velocity in the vertical direction is zero. Hence, Δ p=mu sin θ=1×20×12=10 kg−m/s

A particle of mass $1 k g$ is projected with a velocity of $20 m / s$ at an angle of $30^{\circ}$ with the horizontal. Find the change in the momentum $(k g - m / s)$ of the particle between the points of projection and the highest point. Take $g = 10 m / s^{2}$

The correct option is B $10$
Since the velocity is constant in the horizontal direction, momentum will change only in vertical direction. At the highest point, velocity in the vertical direction is zero.
Hence, $Δ p = m u sin θ = 1 \times 20 \times \frac{1}{2} = 10 k g - m / s$