question_answer

A particle of mass m moving eastward with a velocity V collides with another particle of same mass moving northwards with the same speed V. The two particles coalesce and the new particle moves in NE direction. Calculate magnitude and direction of velocity of new particle.

Answer:

                Refer to Fig. 4(HT). 10. Applying principle of conservation of linear momentum to                        (i) x component of motion \[mV=2mV_{x}^{'}\] \[\therefore \]   \[V_{x}^{'}=V/2\] (ii) y component of motion \[V'=\sqrt{V_{x}^{'2}+V_{y}^{'2}}=\sqrt{{{\left( V/2 \right)}^{2}}+{{\left( V/2 \right)}^{2}}}=\frac{V}{2}\sqrt{2}=V/\sqrt{2}\] \[\tan \theta =\frac{V_{y}^{'}}{V_{x}^{'}}=\frac{V/2}{V/2}=1\]                        \[\therefore \]\[\theta ={{45}^{o}}\]