1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

A
0.2 kg m sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.2 kg m sec1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.1 kg m sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.4 kg m sec1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B −0.2 kg m sec−1From the x−t graph, velocity of particle is given by: v=slope of x-t curve=dxdt ⇒From t=0 to t=2 s, slope of x−t curve is constant, so velocity is, ∴v=slope of x-t curve=4−02−0=2 m/s Just after t=2 s, x=constant v=dxdt=0 m/s ⇒At t=2 s, particle is undergoing a transition in velocity from vi=2 m/s to vf=0 From impulse momentum-theorem on particle: J=ΔP=mvf−mvi ...(i) ∴J=0.1(0−2)=−0.2 kg m sec−1

Suggest Corrections
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program