The attached diagram shows the displacement of the pendulum bob, through an angle theta, as a consequence of the complete absorption of the incident radiation. Now use conservation of mechanical energy.

`Delta K +Delta U=0`

`(K_f-K_i)+(U_f-U_i)=0`

In this case, set the initial potential energy of the pendulum to zero and since the final state of the pendulum comes to a complete stop, `K_f=U_i=0` .

`-K_i+U_f=0`

`U_f=K_i`

`mgh=p_i^2/(2m)`

`mgL(1-cos(theta))=p_i^2/(2m)`

Solve for `theta` .

`eq. (1) :->` `theta=cos^-1(1-P_i^2/(2m^2gL))`

Now use conservation of momentum between the laser and the initial momentum of the pendulum, `p_i` .

`p_i=p_(laser)`

Use the momentum relation for light. Then relate energy to the laser power `P` .

`p_i=E_(laser)/c=(P delta t)/c`

Now substitute for `p_i` in `eq. (1)` .

`theta=cos^-1(1-(P^2(Delta t)^2)/(2m^2c^2gL))`

Substitute numerical values and evaluate `theta` .

`theta=cos^-1(1-((1000 MW)^2(200 ns)^2)/(2(10.0 mg)^2(2.998*10^8 m/s)^2(9.81 m/s^2)(0.0400 m)))=6.10^@`