A particle of mass 100 g is fired with a velocity $20 m s e c^{- 1}$ making an angle of $30^{\circ}$ with the horizontal. When it rises to the highest point of its path then the change in its momentum is

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Solution

The correct option is D

Horizontal momentum remains always constant. So change in vertical momentum $(Δ \to p)$ = Final vertical momentum – Initial vertical momentum $= - m u s i n θ$

$| Δ P | = 0.1 \times 20 \times s i n 30^{\circ} = 1 k g m / s e c$

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A particle of mass 100 g is fired with a velocity 20 m sec−1 making an angle of 30∘ with the horizontal. When it rises to the highest point of its path then the change in its momentum is

The correct option is D Horizontal momentum remains always constant. So change in vertical momentum (Δ →p) = Final vertical momentum – Initial vertical momentum =−mu sin θ |ΔP|=0.1×20×sin 30∘=1 kg m/sec

A particle of mass 100 g is fired with a velocity $20 m s e c^{- 1}$ making an angle of $30^{\circ}$ with the horizontal. When it rises to the highest point of its path then the change in its momentum is

The correct option is D

Horizontal momentum remains always constant. So change in vertical momentum $(Δ \to p)$ = Final vertical momentum – Initial vertical momentum $= - m u s i n θ$

$| Δ P | = 0.1 \times 20 \times s i n 30^{\circ} = 1 k g m / s e c$