Mechanics doubt

I have a doubt in Conservation of Momentum in this problem…


(Source : Anurag Mishra)
@Cooljet123
This is a basic problem only. I’m not so strong in Impulse, Conservation of Momentum chapter :sweat_smile: . So I’m mastering it from the bottom. This is my attempt: Conservation of Momentum is done only in one frame ( inertial ) in both the sides of the equation right? Here the muzzle velocity is referred as the velocity of the Bullet/Rocket with respect to the laucher/barrel… So once the rocket gets out, the velocity of the Man+Launcher changes and shouldn’t (the encircled (in red)) equation be V2=Vo+V1… the whole remaining problem changes with this answer!

Please dont’t mind if its a basic doubt

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Hey, yes you are correct. I can surely confirm it. Momentum conservation is to be applied from ground frame only.
We know muzzle velocity is velocity of rocket wrt rocket launcher AFTER FIRING(it as been given in part A of the question too).
Look at this
V(rocket/rocket launcher) = V(rocket/ground) - V(rocket launcher/ ground AFTER FIRING)
So, V(rocket/ground) = V(rocket/rocket launcher) + V(rocket launcher/ground AFTER FIRING)
plug in the values
V(rocket/ground) = Vo + V1
Same concept is used in trolley problems.
Always remember, even when it isn’t given in question, the muzzle velocity is always the velocity of bullet wrt gun. We conserve wrt ground frame and take velocity of bullet wrt ground AFTER FIRING

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If you still have doubt conceptually, watch fhe following lecture so that you are sure about what I said

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how to solve?

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I THING IT IS ASKING TIME REUIRED TO REACH THE TOP
as at top in limiting case reaction is zero

How to solve

In a hoop, normal just becomes zero at approximately 131degrees when it looses contact with outer wall, not at the top. Root(4gR) velocity is enough to complete the vertical circle in the case of a hoop. After loosing contact at 131degrees, it will take support from inner wall(so, only direction of normal changes after loosing contact of outer wall)
Watch following 5 minute lecture to understand the concept in detail:

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Bhai I know that pls tell me how to solve the question



@Rade_Woosh

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So what is the time taken?

solve the differential put limits of v at starting and at the found angle theta

Nahi horaha

i will solve and upload by 12 pm

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getting wrong ans

Is it c

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Ye how

f=umg
a=ug
v=ugt
Power = fv
Work done = fvt = umgvt
Now substitute ugt = v
Work = mv^2= 10 J

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Here’s a conceptual explanation.
KE finally of block is 5. If we make fbd then friction acts forward on block and backward on plank. So external force has to provide force to make block move forward and also to move plank forward against the friction acting to slow down plank. So equal external force has to be provided to both plank and block. That means same amount of energy is given to them both
. Final energy of block is 5 so that means 5 energy was also used to speed up plank. So ans is 10

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Bc
It free falls untol it makes angle 30 with horizontal.after that only perpendicular component remains.parallel component becomes zero due to impulse tension

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