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

Horizontal momentum remains always constant. So change in vertical momentum (Δ p⃗) = Final vertical momentum – Initial vertical momentum = mu sin θ |Δ P|=0.1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Summary for Time, Height and Range

A particle of...

Question

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

$\sqrt{3}$ $k g m s e c^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1}{2} k g m s e c^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\sqrt{2}$ $k g m s e c^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$1 k g m s e c^{- 1}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$1 k g m s e c^{- 1}$

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

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

Suggest Corrections

Similar questions

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

Q. A projectile of mass

100 g

is fired with a velocity of

20 m / s

making an angle of

30^{0}

with the horizontal.As it rises to the highest point of its path,its momentum changes by

Q. A particle is fired with velocity

u

making angle

θ

with the horizontal. what is the change in velocity when it is at the highest point?

Q. A particle is fired with velocity

u

making angle

θ

with the horizontal. What is the magnitude of change in velocity when it is at the highest point ?

Join BYJU'S Learning Program

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 1 kg m sec−1 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
$1 k g m s e c^{- 1}$

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

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