4)
again its relative velocity
keep the mirror constant
vel of object wrt mirror is 3i+4j+5k-(4i+5j+8k) =-1i-1j-3k
vel of image will be opposite to this
so vel of image will be i+j+3k
am not so sure , but hope someone verifies it
1)A particle moves perpendicularly towards a plane mirror with constant speed of 4cm/s. What is the speed of the image observed by a observer moving with 2cm/s along the same direction? Mirror is also moving with a speed of 10cm/s in the opposite direction.(all speed are with respect to the ground frame)
2)An object O is just about to strike a perfectly reflecting inclined plane of inclination 37°. Its velocity is 5m/s. Find the velocity f the image
3)An elevatr at rest which is at 10th floor os a building is having a plane mirror fixed to its floor. A particle is projected with a speed √2m/s and at 45° with the horizontal. At the very instant of projection the cable of the elevator brakes and the elevator starts free falling. What will be the separation between the particle and its image 0.5 seconds after the instant of projection.
4) A plane mirror is moving with velocity 4i + 5j + 8k. A point object in front of the mirror moves with velocity 3i + 4j + 5k Here k is along the normal to the plane mirror and facing towards the object. he velocity of the image is?
(Bold letters denote the vectors)
5) There is a dust particle on a glass slab of thickness t and refractive index 1.5. When seen from one side of the slab, the dust particle appears at a distance 6cm. From other side it appears at 4cm. find the thickness t of the glass slab
5th one done here:
http://www.goiit.com/posts/list/optics-dust-particle-on-a-glass-slab-1001157.htm
1)
keeping the mirror constant
implies that particle is moving at vel of 14 towards the mirror
and man is moving at 2 towards the mirror
therefore vel of particle wrt man is 14-2=12 which will be the velocity of the image too
hope someone verifies it
4)
again its relative velocity
keep the mirror constant
vel of object wrt mirror is 3i+4j+5k-(4i+5j+8k) =-1i-1j-3k
vel of image will be opposite to this
so vel of image will be i+j+3k
am not so sure , but hope someone verifies it
Answers given in the book ( DC Pandey)
1) 26cm/s
2)4.8i+ 1.4j
3)0.5m
4)3i + 4j + 11k
5)15cm
@ SHANKS SIR
For the 4th one I got the same answer as yours, but its wrong
1st answer wrong according to th book, but I think its correct because am getting the same answer (and its making sense too)
thanks for and the link and help [1]
4) A plane mirror is moving with velocity 4i + 5j + 8k. A point object in front of the mirror moves with velocity 3i + 4j + 5k Here k is along the normal to the plane mirror and facing towards the object. he velocity of the image is?
see velocity of image wrt observer will change only in the direction of the normal from the plane of mirror to the object.
i.e. here only the k component will change . other component will remain same as they have no relation with the mirror's velocity.
So, the vel of image wrt observer (i.e. point object)
= 3i + 4j + (5 + (8-5))k
= 3i+4j+11k
2) similarly find component of motion which is perpendicular to plane of mirror and joins the particle. the relative velocity will be affected only in this direction (similar to Q4.)
if you want derivation for this, try assuming a coordinate system as I have done in solving Q1. but now take two dimensional coordinate systems.