A) 4.0 N
B) \[4/\sqrt{2}N\]
C) \[4\sqrt{2}N\]
D) zero
Correct Answer: C
Solution :
: Resolve\[{{v}_{1}}\]and\[{{v}_{2}}\]along the bat and perpendicular to it along the normal ON Since \[{{v}_{1}}={{v}_{2}},\] components along normal cancel out. components along bat add up. \[\therefore \]change in velocity \[=({{v}_{1}}+{{v}_{2}})\cos 45{}^\circ \] \[=\frac{(v+v)\times 1}{\sqrt{2}}=\frac{2v}{\sqrt{2}}=\sqrt{2}v\] \[\therefore \]change in momentum = mass\[\times \sqrt{2}v=\sqrt{2}\,mv\] time\[=t\] \[\therefore \]Force\[=\frac{\sqrt{2}mv}{t}\] \[=\sqrt{2}\times \left( \frac{200}{1000} \right)\times \frac{40}{2}=4\sqrt{2}N\]You need to login to perform this action.
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