question_answer

Two ice hockey players, suitably padded, collide directly with each other and immediately become entangled. One has a mass of 120 kg and is travelling at 2 m/s, while the other has a mass of 80 kg and is travelling at 4m/s towards the first player. In which direction and at what speed do they travel after they become entangled?

Answer:

Case-I (Before entangling) 1st player 2nd player \[{{M}_{1st\,\,player}}={{m}_{1}}=120\,\,kg\,\,\,{{M}_{2nd\,\,player}}={{m}_{2}}=80\,\,kg\] \[{{u}_{1st\,\,player}}={{u}_{1}}=2m/s\,\,\,{{u}_{2nd\,\,player}}={{u}_{2}}=4\,m/s\] (negative sign as the velocity of the player is opposite) Case-II (After entangling) Total mass after entangling \[={{M}_{1st\,\,player}}+{{M}_{2nd\,\,player}}\] \[={{m}_{1}}+{{m}_{2}}=(120+80)\,kg\] Common velocity of the pair = ?V? m/s Momentum before collision = momentum after collision ..........(A) Momentum before collision \[={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\] \[=120\times 2+80(-4)\] \[=240-320=-80\] ........(1) Momentum after collision \[=({{m}_{1}}+{{m}_{2}})V\] \[=(120+80)V=200V\] ..........(2) \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{}^{10}/{}_{\sqrt{3}}}{{}^{5\sqrt{3}}/{}_{2}}=\frac{10}{3}\times \frac{2}{5\sqrt{3}}=\frac{4}{3}\] substituting (1) and (2) in ((A), we have, \[-80=200V\] \[\Rightarrow V=-\frac{80}{200}=-0.4\,m/s\] We have considered left to right direction as positive. As ?V? obtained is negative, this means that velocity has opposite direction i.e., from right to left. \[\therefore \] the players move from right to left at a velocity of 0.4 m/s.