A particle is projected at angle 60- with speed 10-3-m-s - from the point -A- as shown in the figure- At the same time the wedge is made to move with speed 10-3-m-s towards right as shown in the figure-Then- the time-in s- after which particle will strike the wedge is-

Velocity of particle V_P =10√(3) cos 60^∘î + 10√(3) sin 60^∘ ĵ V_P = 5√(3)î + 15 ĵ Velocity of wedge V_W = 10√(3)î Velocity of particle wrt wedge nbsp; V_PW = ...

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

Byju's Answer

Standard XII

Physics

Relative Velocity in 2D

A particle is...

Question

A particle is projected at angle $60^{\circ}$ with speed $10 \sqrt{3} m/s$ , from the point $‘ A^{'}$ as shown in the figure. At the same time the wedge is made to move with speed $10 \sqrt{3} m/s$ towards right as shown in the figure.Then, the time(in $s$ ) after which particle will strike the wedge is

Open in App

Solution

Velocity of particle
$V_{P} = 10 \sqrt{3} cos 60^{\circ}^i + 10 \sqrt{3} sin 60^{\circ}^j$
$V_{P} = 5 \sqrt{3}^i + 15^j$
Velocity of wedge
$V_{W} = 10 \sqrt{3}^i$
Velocity of particle wrt wedge
$V_{P W} = V_{P} - V_{W} = - 5 \sqrt{3}^i + 15^j$
$| V_{P W} | = \sqrt{25 \times 3 + 225} = \sqrt{300} = 10 \sqrt{3} m/s$
$\Rightarrow tan θ = \frac{15}{5 \sqrt{3}} = \sqrt{3}, θ = 60^{\circ}$
Time after which particle will strike the wedge is
$T = \frac{2 u_{⊥}}{g_{⊥}} = 2 \times \frac{(10 \sqrt{3} sin 30^{\circ})}{(g cos 30^{\circ})}$
$T = \frac{2 \times 10 \sqrt{3} \times \frac{1}{2}}{10 \times \frac{\sqrt{3}}{2}} = 2 sec$

Suggest Corrections

Similar questions

Q. A particle is projected at an angle

60^{o}

with speed

10 \sqrt{3} m / s

, from the point

A,

as shown in the figure. At the same time the wedge is made to move with speed

10 \sqrt{3} m / s

towards right as shown in the figure. Then the time after which particle will strike with wedge is

Q. A particle is projected at and angle 60 with speed

10 \sqrt{3} m / s

. from the point A. as shown in the figure. At the same time the wedge is made to move with speed

10 \sqrt{3}

m/s towards right as shown in the figure. Then the time after which particle will strike with wedge to (g = 10 m/s')

Q. A particle

P

is projected with speed

20 m / s

at angle

37^{o}

from horizontal as shown in figure. At the same instant

Q

starts from same point and moves with uniform acceleration towards right. Path of the particle

P

w.r.t

Q

is straight line. Acceleration of

Q

is:

Q. A particle is projected at an angle of

60^{\circ}

above the horizontal with a speed of

10 m / s

. After some time, the direction of its velocity makes an angle of

30^{\circ}

above the horizontal. The speed of the particle at this instant is

Q. A particle

A

is projected with an initial velocity of

60 m / s

at an angle

30^{\circ}

to the horizontal. At the same time, a second particle

B

is projected in direction as shown in figure with initial speed of

50 m / s

from a point at a distance of

100 m

from

A

. If the particles collide in air, find the time when the particles collide.

Join BYJU'S Learning Program

A particle is projected at angle 60∘ with speed 10√3 m/s , from the point ‘A′ as shown in the figure. At the same time the wedge is made to move with speed 10√3 m/s towards right as shown in the figure.Then, the time(in s) after which particle will strike the wedge is