Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

0.2 kg m s⁻¹

–0.2 kg m s⁻¹

0.1 kg m s⁻¹

–0.4 kg m s⁻¹

Concept: Impulse is defined as the change in momentum ( $\Delta p$ ). $I = \Delta p = m(v_f - v_i)$ [Source 66, 68].
Velocity from Graph: In a position-time ( $x-t$ ) graph, velocity is the slope of the line ( $v = dx/dt$ ) [Source 31].
Calculation:

Before t = 2 s: The particle moves from $x=0$ to $x=2$ m in 2 seconds. $v_i = \frac{2 - 0}{2 - 0} = 1 \text{ m/s}$
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). $v_f = \frac{0 - 2}{4 - 2} = -1 \text{ m/s}$

Impulse: $I = 0.1 \text{ kg} \times (-1 - 1) \text{ m/s}$ $I = 0.1 \times (-2) = -0.2 \text{ kg m s}^{-1}$

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