Engineering Physics help (Frictionless surface with hockey puck)

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The discussion centers on a physics problem involving a hockey puck on a frictionless surface. After a perpendicular blow is delivered to the puck, the trajectory will deviate, potentially forming a 45-degree angle if the vertical and horizontal speeds are equal. The puck's final speed immediately after the blow will be greater than the initial speed due to the addition of vertical velocity from the impact. However, once the blow is delivered, the puck will maintain a constant velocity in both magnitude and direction since no forces act on it afterward. The answers provided for parts A and B are confirmed as correct, while part C clarifies that the puck's motion remains unchanged after the impact.
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This is an example from my engineering physics book which didn't have an answer at the end of the book,
This figure is a snapshot looking down on a frictionless puck moving at uniform velocity from left to right on a level air table. At the position shown, the puck is given a short, sharp hammer blow B in a direction perpendicular to that in which it is initially moving.

0=puck ( Please ignore parentheses in picture they are there to help aligh picture)

-------------------- 0 ------------------
((()))))))))))))))))))))^ Vi[/size]---->
(((((((((())))))))))))))B

A) Show on the figure a trajectory that the puck might follow on the table after the blow is delivered?

My answer - The puck would go exactly 45 degrees to the right (I am not entirely sure - I keep on thinking it is going to go straight)

B) Will the final speed Vi[/size] of the puck (immediately after the blow) be equal to, greater than, or smaller than Vi[/size]? Explain your reasoning?

My answer - equal to because on a frictionless surface the velocity is never going to change

C) How will the velocity of the puck on the frictionless surface behave as time goes by after the blow? That is, will either the magnitude or the direction of the velocity (or both) keep on changing? If so, how?

My answer = ? (baffled on this part)

I am not sure if my answers are right. I need help understanding the third part for sure. Please correct me if I am wrong for my answers a and b.
 
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A) The trajectory will indeed be deviated. However, the angle it will make will be dependend on the "bump". If it is so that the vertical speed is exactly the same as the horizontal speed, then yes, it will give a 45° angle

B) I'm assuming you are familiar with the fact that speed is a vector. Since the bumpt is perpendicular to the traveling speed, the horizontal speed will not change. The vertical speed however, will differ from zero. Hence, V=Vx+Vy, the end velocity will be greater then the begin velocity. Your reasoning is wrong, because while it is true that the velocity will not decrease due to friction, it may increase due to collisions (like the bump)

C) Since there are no forces acting on the puck after the bump, there is no acceleration and the velocity will remain constant, both in magnitude and direction.

Hope it helps :smile:
 


Your answers for parts A and B are correct. For part C, the magnitude and direction of the velocity will remain constant after the blow is delivered. This is because there is no friction on the surface to slow down the puck or change its direction. The hammer blow only changes the direction of the puck, but does not affect its speed. Therefore, the puck will continue to move at a constant velocity in a straight line after the blow.
 
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