Question

The figure shows the position-time (x - t) graph of
one-dimensional motion of a body of mass 0.4 kg.
The magnitude of each impulse is :- (AIEEE-2010) York

Answer 1

U=1 m/s

V= 1 m/s

pi=mf = 0.4*1 = 0.4 Ns

pf = mv = 0.4*(-1) = -0.4 Ns

Impulse = change in momentum = pf-pi

                                                     = -0.4-0.4

                                                    = -0.8 Ns.

Answer 2

Answer:

The magnitude of each impulse is 0.8N-s.

Explanation:

Given the mass of a body,  $m= 0.4 kg$

Slope of the position-time graph gives the velocity of the body.

From the given graph, the initial velocity is,

                              $v_{1} =\frac{\Delta y}{\Delta x} =\frac{2-0}{2-0} =\frac{2}{2} =1m/s$

  the final velocity is,

                              $v_{2} =\frac{\Delta y}{\Delta x} =\frac{0-2}{4-2} =\frac{-2}{2} =-1m/s$

Impulse is defined as change in momentum, and is given by

                               $J=m\Delta v=m( v_{2} -v_{1})$

                                  $=0.4\big( (-1) -1\big)=0.4\times(-2)$

                                  $=-0.8N.s$

The magnitude of impulse is

                              $|\vec J |=0.8N.s$

The magnitude of each impulse is 0.8N-s.