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

Byju's Answer

Standard XII

Physics

Instantaneous Force and Impulse

A body of mas...

Question

A body of mass 0.5 kg is projected under the gravity with a speed of 98 m/s at an angle of $30^{0}$ with the horizontal. the change in momentum ( in magnitude ) of the body when its strike the ground is

98 N-s

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

49 N-s

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

196 N-s

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

24.5 N-s

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

Open in App

Solution

The correct option is B 49 N-s
$49$ $N - s$
Change in momentum $=$ force $\times$ time
$= m g \times t i m e$
$= 0.5 \times 9.8 \times 10$
$= 49 K g m / S$
$(g = 9.8 m / s^{2})$

Suggest Corrections

Similar questions

A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of $30^{\circ}$ with the horizontal. The change in momentum (in magnitude) of the body is

Q. A particle having mass

0.5 k g

is projected under gravity with a speed of

98 m / s e c

at an angle of

60^{\circ}

.The magnitude of the change in momentum

(

in N sec

)

of particle after

10 s e c o n d s

is :

Q. A particle having a mass

0.5 k g

is projected under gravity with a speed of

98 m / s

at an angle of

60^{o}

. The magnitude of the change in momentum in

k g - m / s

of the particle after

10 s

Q. A body of mass

1 k g

is projected from ground at an angle

30^{\circ}

with horizontal on a level ground at a speed

50 m / s

. The magnitude of change in momentum of the body during its flight is

(g = 10 m / s^{2})

Q. A body of mass

1 k g

is projected at an angle

30^{\circ}

with horizontal on a level ground at a speed

50 m / s .

The magnitude of change in momentum of the body during its flight is

(g = 10 m / s^{2})

Join BYJU'S Learning Program

A body of mass 0.5 kg is projected under the gravity with a speed of 98 m/s at an angle of 300 with the horizontal. the change in momentum ( in magnitude ) of the body when its strike the ground is

The correct option is B 49 N-s49 N−sChange in momentum = force × time =mg×time =0.5×9.8×10 =49Kgm/S (g=9.8m/s2)

A body of mass 0.5 kg is projected under the gravity with a speed of 98 m/s at an angle of $30^{0}$ with the horizontal. the change in momentum ( in magnitude ) of the body when its strike the ground is

The correct option is B 49 N-s
$49$ $N - s$
Change in momentum $=$ force $\times$ time
$= m g \times t i m e$
$= 0.5 \times 9.8 \times 10$
$= 49 K g m / S$
$(g = 9.8 m / s^{2})$