# Another Momentum one, sorry

## Homework Statement

A .5kg hockey puck moving at 2 m/sec strikes another puck with a mass of 1kg. The puck then travels at an angle of 60 degrees with a speed of .75m/sec with respect to the motion of the original puck.

a) What is the final speed of the first puck?
b) What is the final direction of the first puck?

m1=.5kg
m2=1kg
v1=?
v2=.75 m/sec

## Homework Equations

p=mv
ΔP1 + ΔP2 = (F1 + F2)(Δt)

## The Attempt at a Solution

m1u1 = (m1+m2)v cos60
m2u2 = (m1+m2)v sin60

This is all I can get. I think I'm right so far lol. I need help badly! This momentum stuff is killing me.

rock.freak667
Homework Helper

## The Attempt at a Solution

m1u1 = (m1+m2)v cos60
m2u2 = (m1+m2)v sin60

This is all I can get. I think I'm right so far lol. I need help badly! This momentum stuff is killing me.

right, so the initial x-direction momentum is m1u1, the initial y-direction momentum is?

But they did not say the masses stick together, so you cannot assume that. Instead assume mass 1's final velocity is v1 (vx1 and vy1 components) and mass 2's is v2, they said v2 is 0.75 m/s at 60°.

So what is the final x-direction momentum and y-direction momentum?

Well I've been working on this question and this is what I came up with. Please take the time to read it and give your input. Thank you very much :):)

m1=.5kg
m2=1kg
v1I=2 m/sec
v2I=0 m/sec
v2F=.75 m/sec

P=mvcos
=(1)(.75)
=.75 kgm/sec

PTotal=m1v1I+m2v2I
PTotal=m1v1F+m2v2F

PTotal=(m1+m2)v
=(.5)(1)(2)
=1kgm/sec

PTotal=m1vF1+m2vF2
1=(.5)vF1 + (1)cos30
1=vF1 + 0.866
.134=vF1 = final speed

rock.freak667
Homework Helper
P=mvcos
=(1)(.75)
=.75 kgm/sec

Why did you neglect the angle?

PTotal=(m1+m2)v
=(.5)(1)(2)
=1kgm/sec

Your answer is correct, but why do you keep assuming that the masses stick together?

PTotal=m1vF1+m2vF2
1=(.5)vF1 + (1)cos30
1=vF1 + 0.866
.134=vF1 = final speed

You have another component to deal with, the vertical component.

Why did you neglect the angle?

Would the angle be 90 degrees which would make the whole thing 0?

Your answer is correct, but why do you keep assuming that the masses stick together?
I just don't know how to do it singly lol.

You have another component to deal with, the vertical component.
Alright I'll work on this

rock.freak667
Homework Helper
Would the angle be 90 degrees which would make the whole thing 0?

The question said at 60°

I just don't know how to do it singly lol.

Just write it out as single components

m1u1+m2u2=m1v1+m2v2

Alright I'll work on this

Remember consider the x and y directions separately at first.

ok i think i got it, hopefully lol:

p=mv

m1u1+m2u2=m1v1+m2v2cos60

(.5)(2)+(1)(0)=.5v1+(1)(.75)cos60

1=.5v1+.375

.625=.5v1

1.25=v1

b) The final direction would just be forward because the value of the speed is positive. Or do you use a formula to find the direction. I'm guessing you use v2=u2+2as, but we don't know acceleration unless it's the gravitational acceleration.

rock.freak667
Homework Helper
ok i think i got it, hopefully lol:

p=mv

m1u1+m2u2=m1v1+m2v2cos60

(.5)(2)+(1)(0)=.5v1+(1)(.75)cos60

1=.5v1+.375

.625=.5v1

1.25=v1

This gives you the horizontal component.

Remember you need to apply conservation of momentum vertically as well.

b) The final direction would just be forward because the value of the speed is positive. Or do you use a formula to find the direction. I'm guessing you use v2=u2+2as, but we don't know acceleration unless it's the gravitational acceleration.

When you get vx and vy for the puck, the direction is just the angle to the horizontal.

Would the vertical component be the same as the first equation except we would use sin60 instead of cos60?

rock.freak667
Homework Helper
Would the vertical component be the same as the first equation except we would use sin60 instead of cos60?

Yes but your initial vertical momentum would be zero.

Alright I got it, thanks alot :D