question_answer

A car of mass 400 kg travelling at 72 kmph crashes into a truck of mass 4000 kg travelling at 9 kmph in the same direction. The car bounces back at a speed of 18 kmph. The speed of the truck after impact is____.

Answer:

Case-I (Before Impact) \[{{M}_{Car}}={{M}_{C}}=400kg\,\,{{M}_{truck}}={{m}_{T}}=4000kg\] \[{{u}_{Car}}={{u}_{C}}=72kg/h\,\,{{u}_{truck}}={{u}_{T}}=9kg/h\] Case-II (After Impact) \[{{M}_{Car}}={{m}_{C}}=400kg\,\,{{M}_{truck}}={{m}_{T}}=4000kg\] \[{{V}_{Car}}={{V}_{C}}=-18kg\,\,{{V}_{T}}=?\] According to law of conservation of linear momentum, Momentum before impact = Momentum after impact ????.. (1) Momentum before impact\[={{m}_{v}}{{u}_{v}}+{{m}_{T}}{{u}_{T}}.....\]? (2) Force and Laws of Motion 25 Momentum after impact \[={{m}_{c}}{{u}_{c}}+{{m}_{T}}{{v}_{T}}.....\] (3) Substituting (2) and (3) in (1), we have, \[{{m}_{c}}{{u}_{c}}+{{m}_{T}}{{v}_{T}}={{m}_{c}}{{v}_{v}}+{{m}_{T}}{{v}_{T}}\] Substituting the above values in (4), we have \[\Rightarrow 400\times 72+4000\times 9=400\times -18+4000\times {{V}_{T}}\] \[\Rightarrow 400[72+10\times 9]=400[-18+18\times {{v}_{T}}]\] \[\Rightarrow 162=-18+10{{v}_{T}}\Rightarrow 162+18=10{{v}_{T}}\] \[\Rightarrow 180=10{{v}_{T}}\Rightarrow {{V}_{T}}=\frac{180}{10}\] \[\Rightarrow {{V}_{T}}=18\,\,kmph\] \[\Rightarrow \] Final velocity of the truck is 18kmph. Therefore, the speed of the truck after impact is 18 kmph.