question_answer

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

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

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

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

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

Correct Answer: B

Solution :

The given situation can be shown as                           Total momentum before impact                 \[={{\mathbf{p}}_{C}}+{{\mathbf{p}}_{L}}=|{{\mathbf{p}}_{C}}+{{\mathbf{p}}_{L}}|\]                 \[=\sqrt{p_{C}^{2}+p_{L}^{2}}\]                 \[=\sqrt{{{(2\times {{10}^{3}}\times 20)}^{2}}+{{(8\times {{10}^{3}}\times 5)}^{2}}}\]                 \[=40\times {{10}^{3}}\sqrt{2}kg\,\,m/s\] 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,                 \[{{10}^{3}}\times 40\sqrt{2}=(2\times {{10}^{3}}+8\times {{10}^{3}})v\] \[\Rightarrow \]               \[v=4\sqrt{2}m/s\]