Home
Class 12
PHYSICS
Two particles having position vectors r(...

Two particles having position vectors `r_(1)=(3hati+5hatj)` metres and `r_(2)=(-5hati-3hatj)` metres are moving with velocities `v_(1)=(4hati+3hatj)m//s` and `v_(2)=(alphahati+7hatj)m//s`. If they collide after 2 seconds, the value of `alpha` is

A

2

B

4

C

6

D

8

Text Solution

Verified by Experts

The correct Answer is:
D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Relative velocity)|5 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (A boat in a river)|3 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Lamni.s theorem)|3 Videos

  • MAGNETISM

    AAKASH SERIES|Exercise ADDITIONAL EXERCISE|22 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise QUESTION FOR DESCRIPTIVE ANSWER|7 Videos

Similar Questions

Explore conceptually related problems

Two particles having position verctors vecr_(1)=(3hati+5hatj) metres and vecr_(2)=(-5hati-3hatj) metres are moving with velocities vecv_(1)=(4hati+3hatj)m//s and vecv_(2)=(alphahati+7hatj)m//s . If they collide after 2 seconds, the value of alpha is

If the position vectors of P and Q are (hati+3hatj-7hatk) and (5hati-2hatj+4hatk) , then |PQ| is

The point having position vectors 2hati+3hatj+4hatk,3hati+4hatj+2hatk and 4hati+2hatj+3hatk are the vertices of

The points with position vectors 10hati+3hatj,12hati-5hatj and ahati+11hatj are collinear, if a is equal to

Two particles of masses m and 2 m has initial velocity vecu_(1)=2hati+3hatj(m)/(s) and vecu_(2)=-4hati+3hatj(m)/(s) Respectivley. These particles have constant acceleration veca_(1)=4hati+3hatj((m)/(s^(2))) and veca_(2)=-4hati-2hatj((m)/(s^(2))) Respectively. Path of the centre of the centre of mass of this two particle system will be:

An object is displaced from position vector r_1=(2hati+3hatj) m to r_2=(4hati+6hatj) m under a force F=(3x^2hati+2yhatj)N Find the work done by this force.

A particular has initial velocity , v=3hati+4hatj and a constant force F=4hati-3hatj acts on it. The path of the particle is

If vec a =hati +3hatj , vecb = 2hati -hatj - hatk and vec c =mhati +7 hatj +3hatk are coplanar then find the value of m.

A particle moves from position r_1=(3hati+2hatj-6hatk) m to position r_2=(14hati+13hatj+9hatk) m under the action of a force (4hati+hatj-3hatk) N , then the work done is

The displacement of a particle from a point having position vector 2hati + 4hatj to another point having position vector 5hatj + 1hatj is

Home
Profile