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

Question

In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t=2 s is : (A) 0.2 kg m s -1 (B) -0.2 kg m s -1 (C) 0.1 kg m s -1 (D) -0.4 kg m s -1

Tardigrade Admin · Accepted Answer

The impulse is given by I = change in momentum =Δ=m Δ v =m (Δ x/Δ t) Given m=0.1 kg , (Δ x/Δ t)=(4.0/2) ms -1 ∴ I=0.1 (4/2)=-0.2 kg - ms -1

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

$0.2 k g m s^{- 1}$

$- 0.2 k g m s^{- 1}$

$0.1 k g m s^{- 1}$

$- 0.4 k g m s^{- 1}$

The impulse is given by
$I =$ change in momentum
$= Δ = m Δ v$
$= m \frac{Δ x}{Δ t}$
Given $m = 0.1 k g, \frac{Δ x}{Δ t} = \frac{4.0}{2} m s^{- 1}$
$∴ I = 0.1 \frac{4}{2} = - 0.2 k g - m s^{- 1}$

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

0.2kgms−1

−0.2kgms−1

0.1kgms−1

−0.4kgms−1