# A pulsed laser fires a `1000 MW` pulse that has a `200 ns` duration at a small object that has a mass equal to `10.0 mg` and is suspended by a fine fiber that is `4.00 cm` long. If the radiation is completely absorbed by the object, what is the maximum angle of deflection of this pendulum?

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^@`