
#1
Apr3007, 06:30 PM

P: 18

1. The problem statement, all variables and given/known data
A 200g puck sliding on ice strikes a barrier at an angle of 53degree and bounces off at an angle of 45degree. The speed of the puck before the bounce was 15m/s and after the bounce it is 12m/s. The time of contact during the bounce is .05seconds A) write these velocities in rectangular components relative to the barrier B) Determine the x and y components of the impulsive force. Which one can be identified as a normal force and which one can be identified as frictional force 2. Relevant equations We were just introduced to impulse last class so i really don't know if I'm doing this right, but i wiki'ed it and see Impulse = Change in Momentum 3. The attempt at a solution A) Pretty sure i did this part right Before: V_x=15cos53 = 9.02723 V_y=15sin53 = 11.9795 After: V_x=12cos45 = 8.48528 V_y=12sin45 = 8.48528 B)This is were I'm kinda lost Formula i used: I=M(v_i  V_f) (M=.2, v_i= velocity initial, v_f= velocity final) I_x=.2(15cos5312cos45)=.541944(.2) = .108389 I_y=.2(15sin5312sin45)=3.49425(.2) = .69885 Guessing the I_y is caused from the normal force of the wall and the I_x is caused from friction. Any help would be appreciated Thanks Tom 



#2
Apr3007, 07:19 PM

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P: 40,883





#3
Apr3007, 07:59 PM

P: 18

Ok now lets see if i get it
A)Before: V_x=15cos53 = 9.02723 (going to the right) V_y=15sin53 = 11.9795 (going down) After: V_x=12cos45 = 8.48528 (going to the right) V_y=12sin45 = 8.48528 (going up) For part B I'm still a little confused because if F_average = M(change in v/change in t) Then Impulse = F_average*Change in t then doesn't that cancel out the t completely? (I'm still working on trying to understand the whole concept, sorry its a little confusing) 



#4
Apr3007, 08:10 PM

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P: 40,883

Impulsive force 



#5
Apr3007, 08:17 PM

P: 18

So this is the same thing i did before but updated with the new signs
B) I_x=.2(8.485289.02723)= .10839 I_y=.2(8.48528(11.9795))= 4.092996 and to solve for the F_average i just do divide by .05 so F_x=.10839/.05 = 2.1678 N F_y=4.092996/.05 = 81.8591 N 



#6
Apr3007, 08:21 PM

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P: 40,883

Good. I didn't check your arithmetic, but it looks reasonable. Now find the force components. Hint: Use the time that was given.




#7
Apr3007, 08:36 PM

P: 18

just updated it with the time
now i got to work on C)what is the direction of the impulsive force with respect to the barrier? I can solve for this by taking the arctan of the two impulse components ? Tan x = opposite / adjacent = 81.8591/ 2.1678 x=88.483 so this means almost all the force is bouncing straight up and a little bit of friction is pushing it back, which seems right to me D) how much energy was lost to heat in the bounce would this be equal to the total impulse ? square root ( .10839^2 + 4.092996^2) = 4.09443 



#8
Apr3007, 08:43 PM

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#9
Apr3007, 08:53 PM

P: 18

so for part D i can just use the difference in Kinetic energy?
(1/2)(.2)(15)^2  (1/2)(.2)(12)^2 = Energy lost due to heat (.1)(225)  (.1)(144)= 8.1 N Well that seems simple if i did it right. And i just want to make sure that on part B when it says "find the components of the impulsive force" Impulsive force is the same thing as F_average? 



#10
Apr3007, 09:05 PM

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#11
Apr3007, 09:07 PM

P: 18

Thanks for all your help



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