• two putty balls of equal masses moving in mutually perpendicular directions with equal speed ,stick together after the collision .if the balls were intially moving with a velocity of 45underroot 2 m/s each feind the velocity of the combined mass after the collision
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Dear Student,

          Let the mass of each ball be 'm' and they are moving with same velocity, v = $45\sqrt{2 } m/s$. After collision they got combined , the mass of combined system= 2m,  let the system moves with velocity v'.
  Now from figure below,
applying conservation law of momentum, in x-axis,
                                                       mv + 0 = 2mv'cosθ​       ...........(i)
       Now in y - axis, 
                                                      mv + 0  = 2mv' sin θ     ...........(ii)

Now adding (i) and (ii) we get,
                                                        2mv = 2mv'(cosθ​ +sinθ)
                                                 $$\begin{gathered} \Rightarrow v\text{'} = \frac{v}{(\cos \theta +\sin \theta )} \\ From figure, \theta ={45}^o \\ \Rightarrow v\text{'} = \frac{v}{(\cos {45}^o+\sin {45}^o)} = \frac{v}{\left({\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}} \right)} \\ \Rightarrow v\text{'} = \frac{45\sqrt{2}\times \sqrt{2}}{2} =45 m/s \\ \mathit{s}\mathit{o}\mathbf{ }\mathit{t}\mathit{h}\mathit{e}\mathbf{ }\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{c}\mathit{o}\mathit{m}\mathit{b}\mathit{i}\mathit{n}\mathit{e}\mathit{d}\mathbf{ }\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}\mathbf{,}\mathbf{ }\mathit{v}\mathbf{\text{'}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{45}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s}\mathbf{.} \end{gathered}$$



Regards