Figure shows the position - time graph of a particle of mass 4 kg, then (a) force on the particle for r lt 0, t gt 4 s, 0 lt t lt 4 s ?(b)impulse at t = 0 and t = 4 s. (consider one - dimensional motion only)

Figure shows the position - time graph of a particle of mass 4 kg, the

1.	Figure shows the position - time graph of a particle of mass 4 kg, then (a) force on the particle for r lt 0, t gt 4 s, 0 lt t lt 4 s ?(b)impulse at t = 0 and t = 4 s. (consider one - dimensional motion only)
Answer» Solution :(a) (i) For `t lt 0`, in the position time GRAPH is at O which means displacement of the particle is zero. i.e., particle is at rest at the origin. Hence force on the particle MUST be zero. (ii) For `0 lt lt 4` s, the position time graph OA has a constant slope. THEREFORE velocity of the particle is constant in this interval i.e., particle has zero acceleration. Hence force on he particle must be zero. (iii) Fro `t GT 4` s, the position time graph AB is parallel to time axis. Therefore, the particle remains at a distance 3 m from the origin, i.e., it is at rest. Hence force on the particle is zero. (b) (i) impulse at t = 0 We know, Impulse = change in linear momentum, Before t = 0 particle is at rest i.e., U = 0. After t = 0, particle has a constant velocity `v=(3)/(4)=0.75 m//s` `therefore` Impulse `= m(v-u)=m(0.75-0)=3kgm//s` (ii) `therefore` Impulse at t = 4 s. Before t = 4s, particle has a constant velocity u = 0.75 m/s After t = 4s, particle is at rest i.e., v = 0 Impulse `= m(v-u)=4(0-0.75)=-3 kg m//s`.