วีดีโอ

thank you sir for this playlist

The increase in kinetic energy during the explosion comes from the internal energy of the shell, which is released in the form of kinetic energy of the fragments.This is a common occurrence in explosive events. Maximizing Velocity: To maximize the velocity of one fragment, we need to minimize the kinetic energy shared by the other two fragments.If the other two fragments have equal velocities, their combined kinetic energy will be minimized for a given total momentum.This allows the third fragment to absorb more of the available kinetic energy, resulting in a higher velocity.

Smooth B@lls: No tangential forces are present, so only the normal components of the velocities are affected. Initial Kinetic Energy of the Striking Ball: K_i = 1/2mv^2 At the moment of maximum deformation: The relative velocity along the normal direction between the balls becomes zero. Why This Happens: At maximum deformation, the balls essentially behave like a single compressed object momentarily, with no relative motion between their centers along the normal direction. This synchronization of velocities is a direct consequence of the balance of forces and conservation laws governing elastic collisions. Velocity Components at Maximum Deformation: At maximum deformation, the two balls move together along the normal direction, sharing a common velocity V_c in that direction. By conservation of momentum along the normal: mv_n1+mv_n2 =(m+m)V_c Decomposing the initial velocity v of the striking ball: v_n1 = vcos(Alpha), v_t1 = vsin(Alpha) Initially, the stationary ball has: v_n2 = 0, v_t2 = 0 Substitute: v_n1 and v_n2 We get, v_c = vcos(Alpha)/2 Their tangential velocities remain unchanged. Kinetic Energy at Maximum Deformation: The total kinetic energy at maximum deformation is the sum of the kinetic energies of both balls: K_maxdef = 1/2(v_c^2+v_t1^2) +1/2(v_c^2+v_t2^2) Potential Energy at Maximum Deformation:The potential energy at maximum deformation is the initial kinetic energy minus the kinetic energy at maximum deformation. U = K_i - K_maxdef

agar tension bhi consider krenge tabhi hanging mass wale ka moment of inertia likhna pdega about o?

Answer not matching sir

The work done by friction is a measure of the energy dissipated from the system.This energy dissipation is a physical reality, independent of the observer's perspective.Therefore, the work done by friction is a frame-invariant quantity.

after the disc pushes off the body M, both the disc and the body M will be moving in the same horizontal direction with the same horizontal velocity. This is because there are no external horizontal forces acting on the system, so the total horizontal momentum must be conserved. By equating the initial and final horizontal momenta, we can determine this common horizontal velocity, V_x. The path of the disc m after leaving body M -: The combination of constant horizontal velocity and increasing vertical velocity downwards results in a curved path, specifically a parabola.

Thank you sir

Sir which force is providing a centripetal acceleration? Mg component?

Why is centrifugal force acting horizontally and not along the rod?

Sir, why can we use just alphaR compared to problem 1.258

pls answer sir is it enough for ADV. I am a JEE 2025 aspirant i have very little amount of time in my hand and almost 70% of my class 12 syllabus is not done properly. losing confidence

sir is this course is upto jee adv level .

what would the ondition have to be for it to not move at all?

Wahhh a shortcut :√ Check out, a=gsin(thetha) ---------- 1 + I --- M R²

I have a total of 18 methods to solve irodov 1.157; one of the best methods involves energy considerations. Potential Energy of the Fallen Part: The fallen part of the chain, of length x, has a mass of λx. Its centre of mass is at a distance x/2 from the table. The potential energy of this part is: Potential energy of the fallen part = (λx)g(x/2) = λgx^2/2 Initial potential energy of the falling part = final kinetic energy of the falling part : λgx^2/2 = (1/2)λxv^2 v = √(gx) Kinetic energy of the falling part = (1/2)λxv^2 = (1/2)λx(gx) = λgx^2/2 Force Exerted by the Falling Part: Work done by the force = change in kinetic energy Fdx = d(λgx^2/2). F = λgx Hence, Force exerted by the fallen part: λgx/2 Force exerted by the falling part: λgx

Velocity of the Center of Mass: V_cm = P_initial/M_total = Vo The initial velocities of the buggies and the man are: Front buggy: V_1,cm = 0 Rear buggy: V_2,cm = 0 Man: u_cm = u - V_cm = u - v_o Let's denote the final velocities of the buggies in the center of the mass frame as V'_1,cm and V`_2,cm . M V'_1,cm + (M + m) V'_2,cm = 0 The relative velocity of the man with respect to the rear buggy remains the same in both frames: V'_2,cm- V'_1,cm = u_cm = u - v_o Solving the Equations, We get V'_2,cm and V'_1,cm. To find the final velocities in the laboratory frame, we add the velocity of the center of mass to the velocities in the center of mass frame.

Jumping in a perpendicular direction means jumping at a right angle to the direction of motion. In the context of the buggies problem, it means jumping sideways, not forward or backward along the tracks. This action allows the men to switch buggies without affecting the overall momentum of the system in the direction of the tracks.

The confusion might arise from the fact that the centrifugal force is often used to explain the outward motion of objects in a rotating frame of reference. However, in the context of the given problem, the work done by the centrifugal force is indirectly accounted for in the work-energy principle. When we calculate the change in kinetic energy, we're implicitly considering the work done by all forces acting on the system, including the centrifugal force.

I think the question is very straightforward. Some people still have some confusion due to both cubes having the same masses. Let's take a lower cube mass as m1 and an upper cube mass as m2. Under compression, the upper cube m2 is shifted from its natural length and comes to equilibrium. Do you know how much compression is needed to bring the m2 cube to its natural length? Yes, that compression amount from which the m2 block is shifted from its natural length. So, you can say, to bring the m2 cube in its natural length, you have to compress more with the same initial compression amount. This means if the initial compression is X_2, then to bring the m2 cube to its natural length, the total compression should be 2*X_2. For the lower cube to bounce up, the spring must extend beyond its natural length. Do you know how much compression is needed to just lift off the lower cube m1? Let's find out. M1g = k*X_1 Where X_1 is that compression that is needed to just lift off the cube m1. So, total compression should be to just lift off the cube m1, 2*X_2 + X_1.

sir humne method 1 mein bouyant force ko kyun nahi liya?

When the spring compresses m1 moves right relative to the center of mass and m2 moves left relative to the center of mass. When the spring is stretched m1 moves left relative to the center of mass and m2 moves right relative to the center of mass. Extension of the Spring at Equilibrium: F_spring = kx Since this force must match the inertial force on m1 due to the center of mass acceleration, we have: kx = m1* a_cm Maximum and Minimum Distances Between the Masses: d_max = L_equilibrium + X_extension = ( lo+x) + x = lo+2x d_min = L_equilibrium - X_compression d_min = (lo+x)-x = lo Too many concepts in this question can't be explained in text.

if i go by TVMF the radius does not get cancel why so?

2:32 its T1 = M1(G-@R) T2 M2(G+@R)

The entire system rests on a frictionless surface, and no external horizontal forces act on the system (apart from the wall, which is initially holding Bar 1 in place). However, after Bar 1 loses contact with the wall, the system becomes isolated in the horizontal direction, meaning the total momentum of the system in the horizontal direction is conserved from that point onward.

m1 and m2 are connected by a spring, which will introduce potential energy into the system as it stretches or compresses. We're asked to find the total energy of this system in the center-of-mass (CM) frame. The spring's potential energy depends on its extension or compression, which is a function of the relative motion of the two masses in the CM frame.However, since the problem doesn't provide specific information about the spring's properties or the initial conditions, we cannot calculate the exact potential energy.Therefore, To fully determine the total energy, we would need more information about the spring, such as its spring constant and the initial displacement from its equilibrium position.

In the equation E' = E + 1/2mv^2, the symbol "v" represents the velocity of the center of mass of the particle system relative to the inertial frame K. To clarify: E: The total energy of the system in the center-of-mass frame (where the center of mass is at rest). E': The total energy of the system in the inertial frame K. m: The total mass of the particle system. The equation E' = E + 1/2mv^2 is valid even if the center of mass is not initially at rest. The key point to remember is that E in this equation represents the total energy of the system in its own center-of-mass frame. Conclusion: The key point to remember is that while the total energy of an isolated system is conserved, the specific distribution of energy into different forms can change between reference frames. This is why, in the equation E' = E + 1/2mv^2, the total energy E' in frame K is different from the total energy E in the center-of-mass frame. The additional term 1/2mv^2accounts for the kinetic energy of the entire system moving with velocity v relative to frame K.

WHY is the direction of velocity not opposite

the center of mass frame provides a unique perspective where the system's kinetic energy is minimized. This principle is fundamental in classical mechanics and has applications in various fields of physics.

but distance covered cant be this what you have compute because the resisting force also be there i.e(bil) so net acceleration hence distance also be pls reply fast sir you are not answering to any query now adays

sir jo force hai BIL vo bhi toh magnetic force hi hai aur usne workdone toh kia hai tabhi ruki hai rod

Greetings sir Sir I wanted to join your course but I am mainly intrested in the pathfinder solutions. So sir are they all the solutions already available in your course or you will upload them with time?

can you explain in detail

sir why cant we do like this: for m2 to move, x=k'm2g/k on m1 evaluating forces, F=km1g+k'x and now substituting x we get F=kg(m1+m2) what is wrong in this method

I think this problem itself is wrong. Firstly, we cannot balance forces on a spherical shell like this because net force on it due to gravity will be zero. You must analyse a portion of the shell using spherical coordinates. Secondly, solids do not exhibit an isotropic pressure in general, and hence there will be a stress in the radial as well as the angular directions inside the earth, and these two stresses will not be equal in general.

sir why did you skip the questions in between😭😭😭

This is an ingenious method !!!

thank you sir . for such a nice and easy explanation

sir how to get to know that whether this self flux would be against or parallel to to flux external..

Apne to F maximum nikala hai . Block ko niche slide hone se bachane keliye F minimum hoga

Sir I have a doubt in this question. Sir we know that V=integral(E.dr) so if E=sigma/4epsilon then V= sigma*R/4epsilon but here the answer is V= sigma*R/2epsilon. Sir why is it that the relation of E and V here fails to give the correct answer? Sir I humbly request you to clear my doubt. Thank you Sir

tried it for 30 mins and even thought of this but didnt applied now i regret

how to solve if all the system lies in a horizontal frame?

Sir what will be the approach to the question if we were given that the velocity varies with s as say As^2 where A is a constant and then we had to find the time taken to hit and were also said to calculate the velocity when it strikes the cylinder

Sir why is the momentum of the system wrt cm zero?

Sir we can directly do it by c equivalent method

Sir in step 3 how you did

Sir ,i found this channel today and i am deeply in love with these videos. My family is not financially stable, so we cant afford books like this, but just because of you, i can solve these books and questions. May god keep you happy and fulfill your dreams. I will definitely promote your channel among my peer group. Love you sir, i cant describe my happiness❤❤❤❤❤❤❤❤❤

does Fl signifies the force due to surface tension?

how do you get ucos theta in 2nd equation?