A ball of mass `m=1kg` strikes a smooth horizontal floor as shown in figure. The impulse exerted on the floor is

Explanation

As the ball strike on the surface on impulsive normal force is exerted on the ball as shown in figure.

This normal force can change only the component `v_(y)`. So in x direction momentum is conserved `(F_("net x")=0)`
`rArr` v' `cos 37^(@) = 5 cos 53^(@)`
`v'=(5xx3xx5)/(5xx4)=15/4` m/sec
So, `v'_(y)=v'sin37^(@)=15/4xx3/5=9/4` m/sec
Impulse = change in linear momentum in y direction
`I=intN.dt=m(v_(y)-(-v'_(y)))=1(4+9/4)`
= 6.25 N-sec.