Home
Class 11
PHYSICS
A particle A moves with velocity (2hati-...

A particle `A` moves with velocity `(2hati-3hatj)m//s` from a point `(4,5m)m`. At the same instant a particle `B`, moving in the same plane with velocity` (4hati+hatj)m//s` passes through a point `C(0,-3)m`. Find the `x`-coordinate (in `m`) of the point where the particles collide.

Text Solution

Verified by Experts

The correct Answer is:
8
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS 1

    DC PANDEY|Exercise MATCH THE COLUMN|10 Videos
  • KINEMATICS

    DC PANDEY|Exercise INTEGER_TYPE|11 Videos
  • LAWS OF MOTION

    DC PANDEY|Exercise Medical entrances gallery|39 Videos

Similar Questions

Explore conceptually related problems

A particle A moves with velocity ( 2 ˆ i − 3 ˆ j ) m / s from a point ( 4 , 5 m ) m . At the same instant a particle B , moving in the same plane with velocity ( 4 ˆ i + ˆ j ) m / s passes through a point C ( 0 , − 3 ) m . Find the x -coordinate (in m ) of the point where the particles collide.

A particle is moving with constant velocity (6hati+8hatj)m//s . It corosses x-axis at point (2,0) . The rate of separation from origin at this moment is :

A particle moving with velocity (2hati-3hatj) m/s collides with a surface at rest in xz–plane as shown in figure and moves with velocity (2hati+2hatj) m/s after collision. Then coefficient of restitution is :

A particle of mass m moving with a velocity (3hati+2hatj)ms^-1 collides with another body of mass M and finally moves with velocity (-2hati+hatj)ms^-1 . Then during the collision

A particle of mass 2.4kg is having a velocity of (3hati+4hatj)m/s at (5,6)m.The angular momentum of the particle about (1,2)m ,in kgm^(2)/s ,is?

A particle of mass 1.0 g moving with a velocity vecv_1 = (3hati - 2hatj) m/s experiences a perfectly inelastic collision with another particle of mass 2.0 g and velocity vecv_2 = (4hati - 6hatj) m/s. Find the magnitude of the velocity vector vecv of the coalesced particles.

A particle moving with a velocity of ( 4hati-hatj ) mis strikes a fixed smooth wall and finally moves with a velocity of ( 3hati + 2hatj ) m/s. The coefficient of restitution between the wall and the particle in the collision will be

A particle is given an initial velocity of vecu=(3 hati+4 hatj) m//s . Acceleration of the particle is veca=(3t^(2) +2 thatj) m//s^(2) . Find the velocity of particle at t=2s.