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 \[{{10}^{3}}\times 40\sqrt{2}=(2\times {{10}^{3}}+8\times {{10}^{3}})v\] \[\Rightarrow \] \[v=4\sqrt{2}\,m/s\]You need to login to perform this action.
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