What is the change in momentum?

[superdave]

Okay, I did this problem, and my answer disagrees with the book.

A 15.0 g rubber bullet hits a wall with a speed of 150 m/s. It bounces straight back with a speed of 120 m/s. what is the change in momentum?

I get:

delta p=(m*v-i) - (m*v-f)
delta p= (15.0g * 150 m/s) - (15.0 g - 120 m/s) = 15.0 g(150m/s - 120 m/s) = 15.0g * 30 m/s
delta p = 450g * m/s = .45 kg * m/s
the book says the answer is 4.05 kg * m/s

Am I wrong?

 

[Libertine]

Yes. Momentum involves direction as well.

 

[superdave]

But that doesn't actually answer my concern. It's the magnitude that disagrees.

Oops, now I see why.

 

[Da-Force]

Okay, I did this problem, and my answer disagrees with the book.

A 15.0 g rubber bullet hits a wall with a speed of 150 m/s. It bounces straight back with a speed of 120 m/s. what is the change in momentum?

I get:

delta p=(m*v-i) - (m*v-f)
delta p= (15.0g * 150 m/s) - (15.0 g - 120 m/s) = 15.0 g(150m/s - 120 m/s) = 15.0g * 30 m/s
delta p = 450g * m/s = .45 kg * m/s
the book says the answer is 4.

​

05 kg * m/s

Am I wrong?

Well, if you look at it vectorally... I usually assume that going away from the wall will be positive and going towards the wall is negative.

So the final momentum minus the initial moment will be your answer.

A shortcut (I prefer) is that the mass is 'constant' for this system so all we see is a change in the velocity.

The change in the velocity is NOT 30 m/s, it is in fact 270 m/s.

So 270 m/s * .015 kg = 4.05 N*s (Newtons per second)

Beware that you might also get this question in many forms with momentum... They try to trick you