# Impulsive force

by tomrule007
Tags: force, impulsive
 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(15cos53-12cos45)=.541944(.2) = .108389 I_y=.2(15sin53-12sin45)=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
Mentor
P: 40,280
 Quote by tomrule007 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
You'll need to know how impulse is defined in terms of force and time. Read this: Impulse of Force
 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
Careful. Signs matter! If the puck has a positive y-component of velocity when it approaches the barrier, what sign must it have after it bounces off?

 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(15cos53-12cos45)=.541944(.2) = .108389 I_y=.2(15sin53-12sin45)=3.49425(.2) = .69885
Two problems: (1) You need to fix your signs, as already mentioned; (2) You've calculated the components of the impulse, but not the force. (Check out the link I gave you.)

 Guessing the I_y is caused from the normal force of the wall and the I_x is caused from friction.
You're on the right track here.
 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)
Mentor
P: 40,280

## Impulsive force

 Quote by tomrule007 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)
That looks good.

 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?
Realize that you're just writing the same equation twice, since Impulse = F_average*Change in t (by definition) and also Impulse = M * change in v (this can be derived from Newton's laws). Pick one version and solve for F!
 P: 18 So this is the same thing i did before but updated with the new signs B) I_x=.2(8.48528-9.02723)= -.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
 Mentor P: 40,280 Good. I didn't check your arithmetic, but it looks reasonable. Now find the force components. Hint: Use the time that was given.
 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
Mentor
P: 40,280
 Quote by tomrule007 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
Looks good.

 Quote by tomrule007 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
Sounds good.

 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
No. Compare the puck's energy before and after the collision.
 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?
Mentor
P: 40,280
 Quote by tomrule007 so for part D i can just use the difference in Kinetic energy?
Yes.

 (1/2)(.2)(15)^2 - (1/2)(.2)(12)^2 = Energy lost due to heat (.1)(225) - (.1)(144)= 8.1 N
Careful with units. Newton's (N) are a measure of force, not energy. (Otherwise: Good!)

 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?
Absolutely.
 P: 18 Thanks for all your help

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