Unit Vector of Blocks Momentum

[rcm431]

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.

 

[Doc Al]

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.

 

[Moetasim]

Hello, it seems like you are working on a problem involving the momentum of a block of ice after being kicked at an angle. I can provide some guidance and clarification on the concept of a unit vector and how it relates to momentum.

A unit vector is a vector with a magnitude of 1 and is used to indicate direction. In this case, the unit vector in the direction of the block's momentum after the kick would be in the same direction as the momentum itself. So, if the block's momentum is in the positive x direction, the unit vector would also be in the positive x direction. This would also be the direction of the block's velocity after the kick.

To find the magnitude of the block's momentum after the kick, you can use the equation P = m*v, where P is momentum, m is mass, and v is velocity. In this case, the mass is given as 0.65 kg and the velocity is 2.5 m/s. So, the magnitude of the block's momentum would be 1.625 kgm/s, as you have calculated.

The z component of the block's momentum after the kick would be zero, as there is no motion in the z direction. The block is only moving in the x and y directions.

I hope this helps with your problem. If you have any further questions, please let me know. Good luck with your calculations!