What is the final speed and direction of the first puck after collision?

[physics(L)10]

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]

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 (v_x1 and v_y1 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?

 

[physics(L)10]

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
v_1I=2 m/sec
v_2I=0 m/sec
v_2F=.75 m/sec

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

P_Total=m₁v_1I+m₂v_2I
P_Total=m₁v_1F+m₂v_2F

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

P_Total=m₁v_F1+m₂v_F2
1=(.5)v_F1 + (1)cos30
1=v_F1 + 0.866
.134=v_F1 = final speed

 

[rock.freak667]

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

Why did you neglect the angle?

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

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

P_Total=m₁v_F1+m₂v_F2
1=(.5)v_F1 + (1)cos30
1=v_F1 + 0.866
.134=v_F1 = final speed

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

 

[physics(L)10]

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]

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.

 

[physics(L)10]

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.

Thank you for any help you give :D

 

[rock.freak667]

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 v_x and v_y for the puck, the direction is just the angle to the horizontal.

 

[physics(L)10]

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

 

[rock.freak667]

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.

 

[physics(L)10]

Alright I got it, thanks a lot :D