Momemuntum of bodies at right angles

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SUMMARY

The discussion focuses on calculating the speed of two colliding particles with masses of 30g and 40g, both traveling at 35m/s at right angles. The momentum conservation equation, MaVa + MbVb = (Ma + Mb)v, is applied to find the common velocity after the collision. The solution involves resolving the momentum in both x and y directions, leading to the resultant velocity calculated using v = √(v_x² + v_y²). This method ensures that both mass and direction are considered in the final speed calculation.

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Homework Statement


Hi
Can someone help with this problem.
2 particles of mass 30g and 40g respectively both travel at a speed of 35m/s in directions at right angles. The 2 particles collide and stick together. Calculate their speed after impact.



Homework Equations


MaVa + MbVb =(Ma + Mb)v
v= common velocity



The Attempt at a Solution


I am not sure how to start. Solving the total momentum before and total momentum after gives the original speed. I then resolved the respective velocities. But that means I got my final value without consideration of the masses! Finally, I resolved the masses into forces, but still didnt get anywhere. Where I am going wrong? Please advice
 
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aurao2003 said:

Homework Equations


MaVa + MbVb =(Ma + Mb)v
v= common velocity

Since they impact at right angles, that means that you have momentum in two directions, x and y.

So for the x-direction you would have

MaVax + MbVbx =(Ma + Mb)vx

and Vbx=0.

Do the same with the momentum in the y-direction and then you know the resultant velocity is given by

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
 
rock.freak667 said:
Since they impact at right angles, that means that you have momentum in two directions, x and y.

So for the x-direction you would have

MaVax + MbVbx =(Ma + Mb)vx

and Vbx=0.

Do the same with the momentum in the y-direction and then you know the resultant velocity is given by

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
Thanks a lot.
 

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