Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t lt 0 , t lt 4s, 0 lt t lt 4s ? (b) impulse at t = 0 and t = 4 s ? ( Consider one-dimensional motion only ) .

Explanation

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 < t < 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 the particle must be zero.
iii) For t > 4 s, the position time graph AB is parallel to time axis. Therefore, the particle remains at a distance 3m 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) = 3` kgm//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