Home
Class 11
PHYSICS
A particle is projected at an angle 60^@...

A particle is projected at an angle `60^@` with speed `10(sqrt3)m//s`, from the point A, as shown in the figure. At the same time the wedge is made to move with speed `10 (sqrt3) m//s` towards right as shown in the figure. Then the time after which particle will strike with wedge is

A

2 s

B

`2(sqrt3) s

C

`4/(sqrt3) s `

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`*rarr_(10sqrt(3) m//s)`
Wedge
`10sqrt(3) sin 60^@ = 15 m//s`

Particle with respect to wedge
`T = (2usin(alpha-beta))/(gcosbeta)`
` =(2(10sqrt(3))*sin(60^@-30^@))/(10cos30^@) = 2s`
.
Promotional Banner

Topper's Solved these Questions

  • PROJECTILE MOTION

    DC PANDEY|Exercise Level - 2 More Than One Correct|8 Videos

  • PROJECTILE MOTION

    DC PANDEY|Exercise Level - 2 Subjective|10 Videos

  • PROJECTILE MOTION

    DC PANDEY|Exercise Level - 1 Subjective|21 Videos

  • MOTION IN A PLANE

    DC PANDEY|Exercise (C )Medical entrances gallery|32 Videos

  • PROPERTIES OF MATTER

    DC PANDEY|Exercise Integer|8 Videos

Similar Questions

Explore conceptually related problems

A particle is projected at an angle 60^(@) with speed 10sqrt3 m/s from the point A as shown in the figure. At the same time the wedge is made to move with speed 10sqrt3 m/s toward right as shown in figure. Find the time after which particle will strike the wedge.

Two particles are projected with speed 4m//s and 3m//s simultaneously from same point as shown in the figure. Then:

Two particles are projected with speed 4 m//s and 3 m//s simultaneously from same point as shown in the figure. Then :

Particle projected from tower fo heigh 10m as shown in figure. Find the time (in sec) after which particle will hit the ground.

A particle is projected from the ground and simultaneously a wedge starts moving towards the right, as shown in the figure. The maximum height of the wedge for which the particle will not hit the wedge is [g=10ms^(-2)]

A ball is projected at an angle 45^(@) as shown with speed 20 sqrt2 m//s at t = 0 . At the same time wedge is made to move with velocity 40 m//s in the direction shown, If the time at which the ball hits the wedge is (sqrtx)/(cos 15^(@)) then value of x is

A particle is projected at an angle 60^(@) with the horizontal with a speed 10m//s . Then, latus rectum is (Take g=10m//s^(2) )

A system is shown in the figure. Block A is moving with 1m//s towards left. Wedge is moving with 1m//s towards right. Then speed of the block B will be :

A particle is projected horizontally with a speed of (40)/(sqrt3)m//s , from some height t = 0 . At what time its velocity makes 60^(@) angle with initial velocity. (g=10m//s^(2))

A particle P is projected with speed 20 m//s at angle 37^(@) from horizontal as shown in figure, At the some 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 : .