A 2.00-kg stone is sliding to the right on a frictionless, horizontal surface at 5.00 m/s when it is suddenly struck by an object that exerts a large horizontal force on it for a short period of time. The graph in ?Fig. E8.13? shows the magnitude of this force as a function of time. (a) What impulse does this force exert on the stone? (b) Just after the force stops acting, find the magnitude and direction of the stone’s velocity if the force acts (i) to the right or (ii) to the left.
Solution 13E Step 1 of 5: The impulse of the net force, denoted by I is defined to be the product of the net force (F) and the time interval(T ) or during for which force acts, That is, I = FT I= F(t 2 t )1.........1 Similarly the change in momentum of a particle during a time interval equals the impulse of the net force that acts on the particle during that interval. I = p f p i Where p and p are initial and final momentum with velocities v and v respectively. f i f i Using P= mv I = mv fmv i I = m(v fv )…i……………..2
