Conservation of Energy Physics Problem, a Physics Problem on Conservation of Momentum?

My experience: To solve physics problems, we must put the conservation of energy in the first place, which is a physical law suitable for the universe. Secondly, there is the conservation of momentum, which is conserved when the resultant force in the direction of momentum is zero or approximately zero.

Force is the object of force, and the motion of the object without force is impulse and momentum equivalent. So, conservation of momentum can be studied.

Total energy loss: 1, collision loss 2, friction produces heat energy. Lost energy = initial kinetic energy of the bullet - kinetic energy at the end of the bullet - kinetic energy at the end of the wooden block.

Solution: Horizontal momentum balance: m bullet * v bullet initial = (m bullet + m block) * v total = >v total = m bullet * v bullet initial (m bullet + m block).

When exercising together:

The kinetic energy lost by the bullet is: m bullet * v bullet initial 2 2-m bullet * v total 2 2 energy balance: the heat energy generated by the friction between the bullet and the wooden block: (m bullet + m wooden block) * v total 2 2-m bullet * v total 2 2

mv0=(m+m)v

Common velocity v=mv0 (m+m).

Kinetic energy lost by the bullet = (1 2) m (v0 -v) = (1 2) mv0 (m +2mm) (m + m).

Heat energy from friction = (1 2) mv0 - (1 2) (m+m) v = (1 2) mv0 m (m+m).

Ratio = (m+m) (m+2m).

ab There is a reason why the velocity of the two balls in the direction of the rope is the same, and the reason is that the rope is restrained. Because the rope is not extendable and the rope is tensioned, the speed of the two ends of the rope along the direction of the rope is the same, and conversely, if the speed of the two balls along the direction of the rope is not the same, it will cause the rope to elongate or the rope to relax. And what is the reason for the equal speed of the two balls?

After the rope is tightened, A, B and the rope system as a whole do horizontal translation, and there is no relative kinematic friend between the three, and there is no sub-velocity of B along the rope. He Yu.

b starts to exercise vb = va

Conservation of momentum: Bu Pai Huai 2

Isn't that because A is located and B is located, and its height is not the same? The higher the place is relative to the lower position, isn't he the potential to close and Wang can be greater? And in the position of the short place, his potential energy is smaller.

So in different positions, the energy is still different.

The answer should be C, because there is resistance in the process of rising and falling, so the mechanical energy must not be conserved, so the mechanical energy must be reduced, whether it is up or down.

In the process of ascent: let h be the maximum height that can be raised, so that mgh+f resistance h=ek, so at point p make 1 2mgh less than 1 2ek, so it has to rise a distance to wait.

In the descent process: mgh-f resists h=ek, so at the p point 1 2ek is less than 1 2mgh, at this time, it has to drop a little more to be equal.

So the answer is only c

The answer is C, mechanical energy includes gravitational potential energy and kinetic energy, rising and falling, both are subject to resistance, so the sum is decreasing, so ab is wrong, the last two answers I suggest you draw a diagram, it is very intuitive and clear.

Incorrect, it does not consider that part of the kinetic energy of the bullet hitting the wooden block is converted into internal energy.

Bullets and a wooden block to observe the state velocity process: mv0 = (m+ma)v1 v1= mv0 (m+ma).

The process of the three is the same velocity: mv0=(m+mA)v1 =(mA+mB+M)v v=mv0 (mA+mB+M).

Sun's (1 2)MV1 2=(1 2)(MA+MB+M)V 2+EP substitutes v1 and v.

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Let the length of the spring group dust be x, and find the derivation of x to obtain x'

x'=0 when the potential energy of the bullet is the largest, so it is assumed to be the middle of the spike front.

EP max=1 2 mV0 2-1 2(mA+mb+m)(mvo) 2

The problem of physical elements (force, motion) is not all vector in the same straight line (force and fortune motion are vectors) can be decomposed by Cartesian coordinates, the principle of selecting the coordinate direction, because there is no friction in the horizontal direction, that is, there is no external force, so the horizontal momentum is conserved, and the vertical momentum is not conserved, so the horizontal and vertical, coordinate decomposition is selected, so that the source solution in the horizontal direction can be conserved by momentum. If the direction of the initial velocity of the bullet is positive, then m bullet x v bullet = m stone x v'Stone - M bullet x v'Bullet level. In the case that the angle of ** back is not known, 300ms is not helpful for solving the above equation.

And v'If the level of the stone and v' bullet is unknown, the above equation cannot be solved, and an additional condition is required, such as the conservation of energy, that is, the conservation of kinetic energy, that is, 1 2 x m bullet x v bullet square = 1 2 x m stone x v'Stone square + 1 2 x m bullet x v'Bullet squared, v'The bullet is known to be 300m s, so v'The stone can be solved, substituting the first spine equation v'The bullet level is solvable, so v'The angle between the bullet and the horizontal can be solved using the inverse cosine function.

This question may seem complicated, but it is actually a fairly basic understanding and application. If you can't think of this decomposition method and the application of momentum conservation, it means that you don't understand enough of the vector decomposition and coordinate decomposition of physics. The mechanical factors in physics can be decomposed in Cartesian coordinates, and the physical factors decomposed in any direction still follow various physical laws in that direction, just to see whether the conditions are met.

The decomposition of this problem satisfies the conservation of momentum in the horizontal direction and the conservation of momentum in the vertical direction, but still satisfies the momentum theorem, from which the impulse of the table against the stone can be calculated.

The first floor is very well analyzed, top!

1) 4mg, downward.

3) Not too late to resist the possibility.

Counter-proof: Assuming that it is possible, let the velocity at this time be v, the force of A on O is downward in the direction of F, the force of B on O is upward in the direction of F, the two forces are equal to each other, and the O axis is not forced, and the equation is:

For A, F+Mg=MV2L

For B, F-2 mg = 2 mV 2 L

The two-formula phase code stove rock subtracts f to obtain 0=v 2 +3mg, which is obviously not valid (because v is argumentatively greater than 0), so the assumption is not valid.