Classic Hit and Stick momentum?

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The discussion focuses on calculating the speed and direction of a blob of clay formed from three colliding balls with different masses and velocities. Participants emphasize the importance of applying the conservation of momentum in both the x and y directions to find the resulting velocity. There is confusion regarding the initial calculations of momentum, with suggestions to clarify the direction of the 45° ball and correctly set up momentum equations. The correct approach involves summing the momenta before the collision and equating them to the combined mass's momentum after the collision. Ultimately, the solution requires dividing the total momentum by the combined mass to find the speed and direction accurately.
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Homework Statement



Three balls of clay have masses m1 = 106 g, m2 = 148 g, m3 = 120 g, and speeds v1 = 1.48 m/s, v2 = 1.19 m/s, v3 = 1.85 m/s. They move in the directions: (1) at 45° from the horizontal, (2) horizontally left, and (3) vertically up. They collide simultaneously and stick together. Calculate the speed and the direction (angle) of the resulting blob of clay.

Homework Equations



p=mv

The Attempt at a Solution



Vx = 106 * 1.48 * cos(45) - 148 * 1.19 = -65.189882 m/s

Vy = 106*1.48sin(45) + 120*1.85 = 332.0309 m/s

V = sqrt(332.0309^2 + [-65.189882]^2) (answer?)

angle = arctan (Vy/Vx)


This isn't right is it,
 
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NoobeAtPhysics said:

Homework Statement



Three balls of clay have masses m1 = 106 g, m2 = 148 g, m3 = 120 g, and speeds v1 = 1.48 m/s, v2 = 1.19 m/s, v3 = 1.85 m/s. They move in the directions: (1) at 45° from the horizontal, (2) horizontally left, and (3) vertically up. They collide simultaneously and stick together. Calculate the speed and the direction (angle) of the resulting blob of clay.

Homework Equations



p=mv

The Attempt at a Solution



Vx = 106 * 1.48 * cos(45) - 148 * 1.19 = -65.189882 m/s

Vy = 106*1.48sin(45) + 120*1.85 = 332.0309 m/s

V = sqrt(332.0309^2 + [-65.189882]^2) (answer?)

angle = arctan (Vy/Vx)


This isn't right is it,
First of all we have to know the direction of the 45° ball. Is it 45° to the left or to the right?

It is not clear what you are doing in calculating vx and vy.

I suggest you write out the equations for momentum before and after the collision and then relate them. How are they related?

AM
 
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No that isn't right, remember that the conservation of momentum applies, use it. Calculate the total momentum in the x direction before the collision, then it them equal to the total momentum in the x direction after the collision, then do the same for the y direction. Then you will have two equations with two unknowns, use algebra to solve for both. What you did was incorrect, you cannot find the total momentum in the x direction and set it equal to the velocity in the x direction. If in doubt, check the units.
 
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So

mv = m1 * v1 + m1*v2 + m2*v3? in the x and y directions then use trig to find speed and direction?
 
NoobeAtPhysics said:
So

mv = m1 * v1 + m1*v2 + m2*v3? in the x and y directions then use trig to find speed and direction?
Yes. What you calculated in the OP (assuming m1 was moving up and to the right, which is not clear) were the x and y momenta, not the x and y speeds. You just needed to divide by the combined mass.
 
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Cool, I got it!
 
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