What is the magnitude of the change in momentum of the racquet ball?

[aszymans]

Homework Statement

Now the racquet ball is moving straight toward the wall at a velocity of vi = 12.4 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -8.2 m/s. The ball exerts the same average force on the ball as before. Mass of ball =.247 kg

Homework Equations

What is the magnitude of the change in momentum of the racquet ball?

The Attempt at a Solution

so I took
m*v(initial)=m*v(final) and got 3.0628 and -2.0254 respectively. so I subtracted the two and got -5.0882. I don't feel like this is correct.

 

[kuruman]

What is incorrect is to say m*v(initial)=m*v(final) because that is simply not true. It is correct to say that the change in momentum Δ(mv)=m*v(final)-m*v(initial), so if the final momentum is negative and the initial momentum is positive, you end up essentially adding two negative numbers and that's that.

 

[PeterO]

Homework Statement

Now the racquet ball is moving straight toward the wall at a velocity of vi = 12.4 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -8.2 m/s. The ball exerts the same average force on the ball as before. Mass of ball =.247 kg

Homework Equations

What is the magnitude of the change in momentum of the racquet ball?

The Attempt at a Solution

so I took
m*v(initial)=m*v(final) and got 3.0628 and -2.0254 respectively. so I subtracted the two and got -5.0882. I don't feel like this is correct.

Many strange "formulae" and calculations, but probably almost the correct answer.

Generally change in a quantity means Final - Initial [which you actually ended up doing!].

Depending which direction you decide to call positive will mean a final answer of -5.etc, like you got in an unusual way, or +5.etc.

The question only asked for the magnitude of the change, so the question of a +ve or -ve answer is irrelevant.