Change in momentum

1. Feb 10, 2009

Victorian91

1. The problem statement, all variables and given/known data
An object of mass m is launched at speed v from point P on horizontal ground at an angle of elevation of 45 degrees, as shown in the picture. When the object reaches point Q the magnitude of the change in momentum of the object is ?

2. Relevant equations
Change in momentum of the object
mv - mu

3. The attempt at a solution
Honestly, I really don't know how to start
All i know is use the formula of the change of momentum
But then the angle 45 degrees I really dont know where to put..
Can somebody help me

Sorry for not attempting
Is because I am a bit blur..
Lastly the options for this question is

A : Zero
B : 1/2 mv
C : 2mv
D : 4mv

Thanks so much..

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2. Feb 10, 2009

Victorian91

Okay i think I might be able to solve it..
Finally I manage to attempt this question..

Initial velocity, v = ( vcos45 )^2 + ( vsin45 )^2 ( using pythagorean theorum ) ( to find the resultant velocity
Therefore initial velocity = v
Hence, initial momentum, p =mv

Final velocity, v is the same in magnitude but opposite direction,
Hence final momentum, p =-mv

Hence, change in momentum,
is -mv -mv

=-2mv

But the question just wants the magnitude,
Hence = 2mv

Can anybody tell me whether my working is correct??
Thanks

3. Feb 10, 2009

gabbagabbahey

Really? Why do you think that?

Also, momentum is a vector:

$$\Delta\vec{p}=\vec{p}_f}-\vec{p}_i=m\vec{v}_f}-m\vec{v}_i$$

What force is causing the momentum to change? Which direction does this force act in? Hence, which component(s) of the momentum change?

4. Feb 10, 2009

Victorian91

The final velocity of the object will be pointing downwards..
Hence, the it will be the same magnitude as the initial velocity but pointing downwards..

That is what I think..