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NCERT Exemplar · Q32

Q.A particle of mass 500 g500\ \text{g} moves in a plane, and its motion is described by two graphs. In the first graph, the xx-coordinate versus time is a straight line through the origin with a positive slope — xx increases uniformly with tt (for example, passing through x=1 mx=1\ \text{m} at t=1 st=1\ \text{s} and x=2 mx=2\ \text{m} at t=2 st=2\ \text{s}). In the second graph, the yy-coordinate versus time is an upward-curving parabola through the origin — yy grows as the square of tt (passing through y=1 my=1\ \text{m} at t=1 st=1\ \text{s} and y=4 my=4\ \text{m} at t=2 st=2\ \text{s}). Find the force (magnitude and direction) acting on the particle.

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Along xx the motion is uniform (straight line), so there is no force in xx. Along yy the position grows as t2t^2, a signature of uniform acceleration ay=2 m s−2a_y=2\ \text{m s}^{-2}. Hence the force is may=1 Nma_y=1\ \text{N} entirely along +y+y.

Concept

Force follows from acceleration by F=ma\mathbf F=m\mathbf a, and acceleration is the curvature (second time-derivative) of each position graph. Here m=500 g=0.5 kgm=500\ \text{g}=0.5\ \text{kg}.

xx-motion

The xx–tt graph is a straight line, so xx is linear in tt: velocity is constant and

ax=0  ⇒  Fx=0.a_x=0\;\Rightarrow\;F_x=0.

yy-motion

The yy–tt graph is a parabola of the form y=12ayt2y=\tfrac12 a_y t^2. Using the point y=4 my=4\ \text{m} at t=2 st=2\ \text{s}:

4=12ay(2)2=2ay  ⇒  ay=2 m s−2.4=\tfrac12 a_y(2)^2=2a_y\;\Rightarrow\;a_y=2\ \text{m s}^{-2}. …

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