Boulder explodes into four pieces

[Glissando]

Homework Statement

A stationary boulder explodes into four pieces. After the explosion, three of the pieces have masses and velocities as follows:

M1: 2.5kg, V: 9.0 m/s due east
M2: 3.5kg, V: 27 m/s due west
M3: 2.0kg, V: 15 m/s due north
M4: 3.8kg, V: ?

Homework Equations

p=mv

The Attempt at a Solution

p_i=0
therefore p_f=0
therefore m1v1+m2v2+m3v3+m4v4 = 0

Here is where I get confused: does the momentum for Mass 1 and 2 "cancel out" since they're in opposite directions? I tried plugging everything in as positive numbers and solving for v4, but that's obviously wrong ):

Thanks for your time and patience!

 

[Delphi51]

Welcome to PF, Glissando.
You have the right idea but you must write separate equations for East-west and North-south. Momentum is conserved independently in each perpendicular direction. With 2 equations, you will find the component of velocity in each direction for the 3.8 kg fragment. You can convert these to polar form to get the magnitude and direction.

 

[Glissando]

Welcome to PF, Glissando.
You have the right idea but you must write separate equations for East-west and North-south. Momentum is conserved independently in each perpendicular direction. With 2 equations, you will find the component of velocity in each direction for the 3.8 kg fragment. You can convert these to polar form to get the magnitude and direction.

Hi Delphi,

Thank you for your quick reply (:!

I'ved tried setting up many equations including p_x = m1v1 + m2v2 = p_y = m3v3 + m4v4

and just through using p_y = m3v3 + m4v4 = 0 to see if I could solve for v4...but to no avail.

I tried this: p_x = m1v1 + m2v2 which resulted in -72kgm/s due west, and since there is 30kgm/s north to counter, I set up a triangle for 72kgm/s due east, 30kgm/s due south and tried pythagorus...which did not give me the correct answer either.

I don't understand what you mean by "polar form" - I don't think I've been taught that before or it's just somewhere in the back of my head ):

Thank you for your help!

 

[gneill]

I tried this: p_x = m1v1 + m2v2 which resulted in -72kgm/s due west, and since there is 30kgm/s north to counter, I set up a triangle for 72kgm/s due east, 30kgm/s due south and tried pythagorus...which did not give me the correct answer either.

When you set up your equations (such as p_x = m1v1 + m2v2), do you mean that you are using the appropriate components of m1*v1 etc., or the magnitude of m1*v1, or something else?

Momentum is a vector quantity, so you must deal with the components separately.

 

[Delphi51]

East-West direction:
mv before = mv after
0 = 9*2.5 - 27*3.5 + 3.8*Vx

Once you find Vx and Vy (north-south component) you have what are called the rectangular components of velocity. The combined velocity is found with the pythagorean formula v = sqrt(Vx² + Vy²) and θ = inverse tan (Vy/Vx). The v and θ are the polar coordinates of the velocity vector.

 

[cupid.callin]

Dont take m4's velocity along Y by your own, you never know where it might fly off. Take velocity of m4 velocity as v_X along east west, v_Y along north south

Find the write the momentum eqn for East-west and North-south and separately find v_X and v_Y and combine them to find mag.

 

[Glissando]

Thank you so much guys,

I figured it out through:

p_x = 0 = 2.5(9) - 3.5(27) + 3.8 (Vx), Vx=18.947368m/s
p_y= 0 = 2.0(1.5)-3.8(Vy), Vy = 7.894736842m/s

Pythagorus, V = 20.52m/s

Thanks once again!

 

[Delphi51]

Good show. The Vy should be negative (no minus sign on the 3.8) but that won't make any difference to your v = 20.52.