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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Jul 2007 02:58:45 IST
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A batsman deflects a ball coming at him with a velocity of 54 km/hr by an angle of 45o, without changing the speed of the ball. What is the magnitude and direction of the impulse imparted to the ball, if the mass of the ball is 0.15kg?
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Will nip in at times to solve problems :)
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1 Jul 2007 06:49:50 IST
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its easy,karthik....use impulse-momentum conservation principle....mv1+Ft=mv2....here ,we take v1 as -vsin 45 degrees..v2 is vsin45degree..it depends on ur choosing the axis..i have chosen normal to the plane at which the ball hits.....m is known to u...subsitute v and find Ft....which is nothing but the impulse imparted....hope u got it..cheers!!
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1 Jul 2007 11:55:28 IST
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well in this case you have to you have to find the change in momentum of both the axis x and Y . in x... change in momentum =
mvcos45-(-mv)=mvcos45+mv similarly in y dirn change in momentum. =mvsin45....conbine them by vector addition and you will get your answer.
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1 Jul 2007 15:17:58 IST
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That is not the answer. The change in momentum, is 2mvcos22.5.
I got it as 2mvcos45.
In the answer, its given that the direction of impulse is along the angle bisector. I did not understand that part. So someone explain.
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well see.................
the angle of deflection is 45deg
now resolve the initial and final vel in rectangular components
for this u'll have to draw a perpendicular bisec of angle of deflection(as y axis)
u(ini vel)=vcos22.5 (along xaxis) -vsin22.5 along perp(yaxis)
final vel =vcos22.5 and vsin22.5
impulse =change in momentum
note that there is no change in momentum along x axis
along yaxis-------------
=mvsin22.5-(-mvsin22.5)
=2mvsin22.5 (along perp bisec)
hope u understood!!!!!!
thank u !!!!!!!!
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1 Jul 2007 17:04:09 IST
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Duh uh you made some blundr with the soln I gave you......AS per my soln. If you find the change in momentum about the axis... youvil get mgcos45+mg in x...and mgsin45..in y..Now add vectoriaaly
under root mg+mgcos45 whole square+mgsin45 whole square.
which Does not give 2mgcos45....Check your calculations..and youwill get your answer... I have calculated myself... then youse tan theta= bsin theta/a+bcos theta to find out the angle .....My soln was absolutely right. I hope i have been able to convince, n if not....tell me n I will try harder...Ok.. take care  |
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2 Jul 2007 02:57:00 IST
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The angle of deviation is  /4. See the diagram attached. The angle between the two momenta vectors is 3 /4. (Angle between two vectors is always measured as the smaller of the two angles when they are placed tail to tail).
Now,
Impulse, I =  p =  (p2+ p2- 2 x p x p x cos3/4) = 2psin(3/8) = 2psin( - 3/8) = 2psin(/8) = 2psin22.50 = 2mvsin22.50
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2 Jul 2007 11:03:02 IST
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Metthod given by elessar is correct but smaller angle angle b/w two vector is 45 .... you can checkit by shifting vectors parallely and join their tails.......
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2 Jul 2007 13:27:21 IST
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How is it 45?? The other angle is 360 - 135 which can't be the smaller angle
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2 Jul 2007 13:36:53 IST
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As we have to take change in vector so replace vector p1
with vector -p1 ....
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2 Jul 2007 13:48:35 IST
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I have used the formula for subtraction of vectors directly. One doesn't need to reverse any vector in that.
If you reverse the vector, then you'll be using the addition formula
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2 Jul 2007 13:54:48 IST
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ok u r correct....
but y not r we considering angle b/w p2 and -p1.....
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2 Jul 2007 14:36:08 IST
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Nice solution elessar, but I have one doubt. Why is the angle between the momentum vectors taken as 135 degrees and not as 45 degrees?
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2 Jul 2007 14:39:31 IST
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as far as i think
method by elessar_iitkgp is correct(but pie/8 is 22.5 not 15)
@ maverick83
we can take the ang b/w p2 ans -p1 too.ans will be the same
if ang b/w i vec and another is theta then b/w the -ve vec and other is 180-theta .makes no difference.
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2 Jul 2007 14:41:27 IST
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