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The momentum of a body in two perpendicular direction at any time ‘t’ are given by $P_{x} = 2 t^{2} + 6$ and $P_{y} = \frac{3 t^{2}}{2} + 3$ . The force acting on the body at t = 2 sec is

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5 units

2 units

10 units

15 units

answer is C.

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Detailed Solution

F= $\frac{d p}{d t}$

$F_{X} = \frac{d p_{x}}{d t} = 4 t = 4 \times 2 = 8 N a n d F_{Y} = \frac{d p_{y}}{d t} = 3 t = 3 \times 2 = 6 N \Rightarrow F = \sqrt{{F^{2}}_{X} + {F_{Y}}^{2}} = \sqrt{8^{2} + 6^{2}} = 10 N$

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