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