question_answer

A car weighing \[\text{2}\,\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\,\text{kg}\]and moving at \[20\text{ }m/s\]along a main road collides with a lorry of mass \[8\times {{10}^{3}}\,kg\]which emerges at \[5\text{ }m/s\]from a cross road at right angle to the main road. If the two vehicles lock, what will be their velocity after the collision?

A) \[4/\sqrt{2}\,m/s,\,{{45}^{o}}\] with cross road

B) \[4/\sqrt{2}\,m/s,\,{{45}^{o}}\] with main road

C) \[4/\sqrt{2}\,m/s,\,{{60}^{o}}\] with cross road

D)  \[4/\sqrt{2}\,m/s,\,{{60}^{o}}\]with main road

Correct Answer: B

Solution :

 The given situation can be shown as Total momentum before impact \[={{p}_{C}}+{{p}_{L}}=|{{p}_{C}}+{{p}_{L}}|=\sqrt{p_{C}^{2}+p_{L}^{2}}\] \[=\sqrt{{{(2\times {{10}^{3}}\times 20)}^{2}}+{{(8\times {{10}^{3}}\times 5)}^{2}}}\] \[=40\times {{10}^{3}}\sqrt{2}\,kg\,m/s\] Direction of momentum with main road,               \[\tan \theta =\frac{{{p}_{L}}}{{{p}_{C}}}\] \[=\frac{8\times {{10}^{3}}\times 5}{40\times {{10}^{3}}}\] \[=1\] \[\Rightarrow \] \[\theta ={{45}^{o}}\] By the law of conservation of linear momentum \[{{10}^{3}}\times 40\sqrt{2}=(2\times {{10}^{3}}+8\times {{10}^{3}})v\] \[\Rightarrow \] \[v=4\sqrt{2}\,m/s\]