What is the magnitude of the impulse from the spike?

[andrewn]

Homework Statement

A ball is traveling horizontally over a volleyball net when a player spikes it, driving it straight down to the ground. The ball's mass is .22kg, its speed before being hit is 6.4 m/s and its speed immediately after the spike is 21m/s. What is the magnitude of the impulse from the spike? I have two attempts at the solution but I was mainly wondering which attempt(s) is correct.

Homework Equations

impulse=change in momentum= (mass)(final velocity)-(mass)(initial velocity)

The Attempt at a Solution

J=.22(21)-.22(6.4)=3.212kg(m/s)
I don't know if that is correct because a way I was given to do this was using coordinates:
.22(0,21)-.22(6.4,0)=(0,4.62)-(1.408,0)=(-1.408,4.62)kg(m/s)

 

[rootX]

Initially, it was traveling horizontally

and after the applied force, the direction changed by 90 degrees.

and impulse is a vector quantity, so you need to take of directions also,.

so your expression would be:
J=.22(21[down])-.22(6.4

)​

 

[Doc Al]

J=.22(21)-.22(6.4)=3.212kg(m/s)
I don't know if that is correct because a way I was given to do this was using coordinates:
.22(0,21)-.22(6.4,0)=(0,4.62)-(1.408,0)=(-1.408,4.62)kg(m/s)

The first way is not correct, since it ignores the fact that momentum (and velocity) is a vector--direction counts. The second method, using components, is correct. (Except for a minus sign, since the final velocity is 21 m/s down.)

To complete the solution, you need to find the magnitude of the impulse, not just the components.