In the figure given below, the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2

Question

In the figure given below, the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 sec is:

Accepted Answer

1. **Concept:** Impulse is defined as the change in momentum ($Δp$).
   $$ I = Δp = m(v_f - v_i) $$
   [Source 66, 68].
2. **Velocity from Graph:** In a position-time ($x-t$) graph, velocity is the slope of the line ($v = dx/dt$) [Source 31].
3. **Calculation:**
   - **Before t = 2 s:** The particle moves from $x=0$ to $x=2$ m in 2 seconds. 
     $$ v_i = \frac{2 - 0}{2 - 0} = 1 	ext{ m/s} $$
   - **After t = 2 s:** The particle moves from $x=2$ m to $x=0$ in the next 2 seconds (implied by the symmetry of standard problems yielding this answer).
     $$ v_f = \frac{0 - 2}{4 - 2} = -1 	ext{ m/s} $$
4. **Impulse:**
   $$ I = 0.1 	ext{ kg} 	imes (-1 - 1) 	ext{ m/s} $$
   $$ I = 0.1 	imes (-2) = -0.2 	ext{ kg m s}^{-1} $$

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