What Is the Unit Vector in the Block's Momentum After a Kick?

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SUMMARY

The discussion centers on calculating the unit vector of a block's momentum after a kick. A 0.65 kg block of ice, initially sliding at 2.5 m/s, is kicked perpendicular to its path, resulting in a new direction at an angle of 28 degrees. The x-component of the block's momentum after the kick is determined to be 1.625 kg·m/s. The key takeaway is that the unit vector can be derived from the angle provided, simplifying the calculation of momentum without needing extensive knowledge of momentum principles.

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rcm431
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Hello everyone, new to the forums but I have used the information here for help in the past. Having trouble with a specific part of a problem.1. A 0.65 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.0025 seconds. The block eventually slides at an angle of 28 degrees from its original direction. The overhead view shown in the diagram is approximately to scale. The arrow represents the average force your toe applies briefly to the block of ice.

A. What is the unit vector in the direction of the block's momentum after the kick?

B. What is the magnitude of the block's momentum after the kick?

C. What is the z component of the blocks momentum after the kick?

2.p^^\-> = abs(p^^\->) p^^\^...p^^\-> = m*v^^\->
3. The ice is sliding as if you are looking down on it (the y-axis), and the kick being in the positive z direction, and the ice sliding in the positive x direction. I know if I find the unit vector of the block's momentum it will be easy to calculate the rest of the problem, part A is the part I am having trouble with.

I calculated the x-component of the block's momentum after the kick which is 1.625 kgm/s


Any feedback would be greatly appreciated.
 
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rcm431 said:
The block eventually slides at an angle of 28 degrees from its original direction.
They just want a unit vector in this direction. (This part of the problem doesn't require any knowledge of momentum, since they give you the direction.)

If you need more help, please attach the diagram.
 

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