Ncert Exercises & Solutions

Q. 32 Figure shows (x, t), (y, t) diagram of a particle moving in 2-dimensions.

If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.

Hint: Velocity, v=

\frac{d x}{d t}

and Acceleration, a =

\frac{d v}{d t}

Step 1: Find acceeration along x-axis.

Clearly, from diagram (a), the variation can be related as

$\begin{array}{l} x = t \Rightarrow \frac{dx}{dt} = 1 m / s \\ a_{x} = 0 \end{array}$

Step 2: Find acceleration along the y-axis.

$From diagram (b) \begin{array}{l} y = t^{2} \\ \Rightarrow \frac{dy}{dt} = 2 t or a_{y} = \frac{d^{2} y}{{dt}^{2}} = 2 m / s^{2} \\ Step 3 : Find net force \\ Hence, F_{y} = {ma}_{y} = 500 \times 10^{- 3} \times 2 = 1 N (∵ m = 500 g) \\ F_{x} = {ma}_{x} = 0 \\ Hence, net force, F = \sqrt{F_{x}^{2} + F_{y}^{2}} = F_{y} = 1 N (along y - axis) \end{array}$

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