- #1

unscientific

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## Homework Statement

The problem is given in the picture: (please refer to the 2 pictures below, thank you!)

I'm having trouble with the last part...

## Homework Equations

Conservation of momentum in both x and y - directions.

## The Attempt at a Solution

From symmetry it can be seen that ball D would be directly below ball C, with ball C moving at speed of (2/5)sqrt(3)*u at an angle of 30 degrees below horizontal.

__Here's my attempt using the x-y direction conservation of momentum__

Taking downwards and leftwards as positive,

**Horizontally**:

There would be no transfer in horizontal momentum to ball D, hence ball C still would have a horizontal velocity of: (3/5)u

**Vertically**(downwards as positive):

sqrt(3)/5 * u = v

_{c}cosθ + v

_{D}

**Energy equation**:

(12/25)u

^{2}= (v

_{c})

^{2}+ (v

_{D})

^{2}

That leaves me with 2 equations with 3 unknowns, v

_{c}, v

_{D}and θ.

I tried using a

__new approach__, but would like to ask if this obeys the laws of physics:

**Horizontal speed of c**: unchanged = (3/5)u

**Considering collision "vertically"**:

can i simply do the momentum equation by solely considering only the vertical velocities?

(1/5)sqrt(3)* u = V

_{c,y}+ V

_{D}

**Again only considering energy in the vertical direction: ( i know this is wrong as energy is not a vector!)**:

(9/25)u

^{2}= (v

_{c,y})

^{2}+ (v

_{D})

^{2}

This allows me to solve the problem, but is this the right approach?

Here's a quick, not-so-rigorous proof: vx

^{2}+ vy

^{2}= v'

^{2}, multiplying (1/2)m throughout.... gives us the split energy equation!)

#### Attachments

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