26 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Have you tried it
Mv = mv'
mgh = 1/2 m(v')^2 + 1/2 M(v)^2
Wait we are looking for distance
Huh is the question wrong ?
nah
D?
Oh wait yes it’s D
yep , send soln @itsav23
@hardcoreisdead
There’s one capital M in the denominator it’s looking like small m ig
???
Can you explain the last step I think I made a mistake in that
Direct formula lagaya hai (Mass/total mass) x displacement of other body
com ka displacement zero rakhna hai na to wedge ko right side itna amount move krna pdega
Oh
Yeah havent you guys studied this short trick method?
Nah I was trying to do it from conservation of momentum
Yeah that would work as well
Long method though
Idk what mistake I made
Yeah but I think I got your method too
Neat way
Yeah this works for the boat one and all types of problem wherever the disp of com in x direction is zero
xcm = mx1+Mx2/(m+M)=0
ie mx1=-Mx2
mx1/M=-x2
x1 kaise aayega
Bhai length or angle islie to dia hai
Jo tumhara block hai wo along the incline fislega niche koi friction wriction nhi hai to wedge ke respect me to uska displacement x component hi to ayega disp ka
Here is the full proof i think it should be clear now
Yep, use this formula. Just make sure that you're applying com in the direction where no force is acting here horizontal direction so hcottheta
tysm
+solved @Real potato000
+solved @Real potato
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