A mass of 2.9 kg suspended from a string of length 50 cm is at rest. Another body of mass 100 g which is moving horizontally with a velocity of 150 m/s stricks to it. What is the tension in the string when it makes an angle of 60°  with the vertical ?

Applying the law of conservation of linear momentum to the collision , the velocity V of the composite mass after collision will be given by 
$0.1\times 150=\left(2.9+0.1\right)\mathrm{V}$
or
$\mathrm{V}=5\mathrm{m}/\mathrm{s}\left(\mathrm{M}=2.9+0.1=3\mathrm{kg}\right)$
Now to calculate the velocity v of composite body when it makes an angle of 60° with the vertical we apply law of conservation of mechanical energy between L and P, i.e  
$\frac{1}{2}{\mathrm{MV}}^2=\frac{1}{2}{\mathrm{Mv}}^2+\mathrm{Mgh}$
i.e,. ${\mathrm{v}}^2={\mathrm{V}}^2-2\mathrm{gL}\left(1-\mathrm{cos\theta }\right)\left[\mathrm{as}\mathrm{h}=\mathrm{L}\left(1-\mathrm{cos\theta }\right)\right]$
or ${\mathrm{v}}^2=5^2-2\times 9.8\times 2\left[1-\frac{1}{2}\right]$
or ${\mathrm{u}}^2=20.1{\left(\mathrm{m}/\mathrm{s}\right)}^2$
Now as for circular motion of p
$\left({\mathrm{Mv}}^2/\mathrm{L}\right)=\mathrm{T}-\mathrm{Mgcos\theta }\left[\mathrm{as}\mathrm{r}=\mathrm{L}\right]$
So T $=\frac{3\times \left(20.1\right)}{\left(1/2\right)}+3\times 9.8\times \frac{1}{2}=120.6+14.7=135.3\mathrm{N}$