# Momentum problem.

## Homework Statement

A ball with a mass of 100g is moving with a velocity of 1 m/s . It collides into a wall, from which it bounces off with the same speed and the same angle. Calculate the change in momentum if
a) it falls perpendicular to the wall (correct answer 0.2kg*m/s)
b) it falls with an angle of 30° (correct answer 0.17 kg*m/s)
I would aprecciate if someone could help me solve this. We only started covering this topic today and I didn't quite understand it.

P.S if the writting seems strange/incorret I am very sorry it's hard to translate from Slovenian

P=m*v
P=m*v*30cos

## The Attempt at a Solution

for a)I have tried using this equation
P=m*v
which got me
P=0.1kg*1m/s=0.1Ns (I am supposed to get 0.2Ns)
for b) I tried
P=0.1kg*1m/s*30cos=0.086Ns.(I am supposed to get 0.17 Ns)

I think you are forgetting about the initial momentum of the ball. The change in momentum is the final momentum minus the initial momentum.

• 1 person
I believe the inital momentum is 0 the since it doesn't give me the Time that the ball traveled, but correct me if I'm wrong

Momentum doesn't care about time, it is simply mass times velocity.

• 1 person
It helps to draw diagrams anytime you are struggling wrapping your head around a concept.

In this case you have an initial momentum before the ball hits the wall, because the ball has a mass and it is moving.

Note that Momentum is ALWAYS conserved.

Edit: Did away with potentially confusing, poorly worded, section.

Last edited:
• 1 person
So what is the correct way to get the answer I still don't get it sorry

If you cannot deduce what the change in momentum is, then you need to refer to your text book, plus it was told to you in the first response.

Chestermiller
Mentor
Momentum is a vector quantity, so the change in momentum is equal to the difference between the final momentum (vector) and the initial momentum (vector). The velocity of the ball changes direction when it hits the wall, so its momentum vector also changes direction. You need to take this into account in part 1. In part 2, the one component of the momentum vector changes direction, but not the other component. So the change in momentum is only the difference between the one component that changes direction.

Chet

• 1 person
Thanks everyone after reading all of this through again I think I got it.
so if I understand this now;
a)P=Pf-Pi=((0.1kg)*(-1m/s))-((0.1kg)*(1m/s))=(-0.1Ns)-(0.1Ns)= the change in momentum is 0.2Ns?
and
b) is basically the same just diffrenct numbers?

Chestermiller
Mentor
Thanks everyone after reading all of this through again I think I got it.
so if I understand this now;
a)P=Pf-Pi=((0.1kg)*(-1m/s))-((0.1kg)*(1m/s))=(-0.1Ns)-(0.1Ns)= the change in momentum is 0.2Ns?
and b is basically the same with diffrenct numbers?
In part b, you need to do it in component form, determining the change in momentum in each of the coordinate directions.

• 1 person
Yeah I did that and I got this equation
P=m*v*30cos

Chestermiller
Mentor
Yeah I did that and I got this equation
P=m*v*30cos
Great. So now get the change in P.

Chet