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

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 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bb66ab248068cd28455e65efcc0190ef-.png" /> (A) 0.2 kg m sec-1 (B) -0.2 kg m sec-1 (C) 0.1 kg m sec-1 (D) -0.4 kg m sec-1

Tardigrade Admin · Accepted Answer

Velocity between t=0 and t=2 sec ⇒ v1=(dx/dt) =(4/2)=2 m/s Velocity at t=2 sec, vf=0 Impulse = Change in momentum =m(vf-vt) 0.1 (0-2)=-0.2 kg m sec-1

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

$0.2 k g m se c^{- 1}$

$- 0.2 k g m se c^{- 1}$

$0.1 k g m se c^{- 1}$

$- 0.4 k g m se c^{- 1}$

Velocity between $t = 0$ and $t = 2 sec$

$\Rightarrow v_{1} = \frac{d x}{d t}$

$= \frac{4}{2} = 2 m / s$

Velocity at $t = 2 sec, v_{f} = 0$

Impulse = Change in momentum $= m (v_{f} - v_{t})$

$0.1 (0 - 2) = - 0.2 k g m se c^{- 1}$

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

0.2kgmsec−1

−0.2kgmsec−1

0.1kgmsec−1

−0.4kgmsec−1