Question

 A small block of mass m and concave mirror of radius R fitted with a stand, lie on a smooth, horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t=0 towards the mirror so that it starts moving towards the mirror at speed V and collides with it. The collision is perfectly elastic.  

Find the velocity of the image 

1) at a time t< d/V 

(a)R² V/[2(d-Vt)-R]² 

(b) R² V/[2(d+ Vt)-R]² 

(c)R²V/[2(d-Vt)-R]² 

(d) RV [1+ R²/[2(Vt-d)-R]² 

 2) at a time t>d/v 

(a)R² V/[2(d-Vt)-R]² 

(b) R² V/[2(d+ Vt)-R]² 

(c)R²V/[2(d-Vt)-R]² 

(d) RV[1+ R²/[2(Vt-d)-R] 

Answer 1

 (1) t < (d/V) 

u = - (d- Vt) 

f = -R/2 

We know that VI/m = -m² V_O/m 

Here,m = f/ (f-u) 

m = (-R/2)/[(-R/2)+(d-Vt)] 

m = (-R)/[2(d-Vt)-R] 

Velocity of image= V_I/m = -m² V_O/m 

V_I= R² V/[2(d-vt)-R]² 

(2) t > (d/v) 

Block will collide with mirror assembly after time T = d/V. From conservation of linear momentum, block and mirror assembly will exchange their momentum i.e., block will stop and mirror starts moving with velocity V. 

 U = -V [t-(d/V)] 

Also,                       m = f/(f- u) = [-R/2]/[(-R/2)+V(t-(d/V))] 

We know that V_I/m = -m²V_O/m VI –V_m = -m2V_O/m 

Let us assume rightward direction as positive V_I –V_m =-m²(-V)        or       VI = (1+m²)V 

V_I=V[1+{R²/2(Vt-d)-R²}]