The position (x) `rarr` time (t) graph of one dimensional motion of a body of mass 0.4 kg. What is the magnitude of impulse of force

Options

A. 1.6 Ns
B. 0.8 Ns
C. 0.4 Ns
D. 0.2 Ns

Answer

A

Explanation

(A) 1.6 Ns
Initial velocity of body is `V_(1) `
=Slope of graph x-t in first two second
`=(4-0)/( 2-0) =2 m//s`
Final l velocity of body is `v_(2)`
slope of `x-t` graph in last two second
`=(4-0)/( 0-4) = -2 m//s`
Impulse of force `|Delta p| = |m (v_(2) - v_(1) )|`
`=1.6 Ns`