The wood block with mass m is placed on a smooth horizontal plane, and the bullet with mass m is fired in the horizontal direction at velocity V0 and finally remains in the wood block and moves with the wood block in V A,FL=1/2Mv2 B.Fs=1/2mv2 C.Fs=1/2mv02-1/2(M+m)v2 D.F(L+s)=1/2mv02-1/2mv2 The wood block with mass m is placed on a smooth horizontal plane, and the bullet with mass m is fired in the horizontal direction at velocity V0 and finally remains in the wood block and moves with the wood block in V. It is known that when the bullet is stationary relative to the wood block, the wood block advances L, and the depth of the bullet entering the wood block is s. if the resistance F of the wood block to the bullet is constant, then

The wood block with mass m is placed on a smooth horizontal plane, and the bullet with mass m is fired in the horizontal direction at velocity V0 and finally remains in the wood block and moves with the wood block in V A,FL=1/2Mv2 B.Fs=1/2mv2 C.Fs=1/2mv02-1/2(M+m)v2 D.F(L+s)=1/2mv02-1/2mv2 The wood block with mass m is placed on a smooth horizontal plane, and the bullet with mass m is fired in the horizontal direction at velocity V0 and finally remains in the wood block and moves with the wood block in V. It is known that when the bullet is stationary relative to the wood block, the wood block advances L, and the depth of the bullet entering the wood block is s. if the resistance F of the wood block to the bullet is constant, then

Choose a, C, D
Momentum conservation: MV0 = (M + m) V
Total kinetic energy change: FS = MV0 ^ 2 / 2 - (M + m) V ^ 2 / 2 = MV0 ^ 2 / 2 - (MV0) ^ 2 / [2 (M + m)] = mmv0 ^ 2 / [2 (M + m)]
Among them, the kinetic energy change of wood block: FL = MV ^ 2 / 2
Bullet kinetic energy change: F (L + s) = MV0 ^ 2 / 2-mv ^ 2 / 2

Energy of a thin disk with mass m, radius R and angular velocity W?

When rotating around the axis perpendicular to the disc center and the disc surface, the moment of inertia is
J=(1/2)mr^2
Rotational kinetic energy EK = (1 / 2) JW ^ 2 = (1 / 4) m (RW) ^ 2

The radius r disk rotates at an angular velocity W around the center O. there are two points a and B at different positions on the edge. Obviously, the velocity directions of a and B are different But I assume that the moving reference system is established with point a as the center. The speed of the moving reference system is RW, and point B is stationary relative to point A. why is the speed direction the same as point a? What's wrong with me? Thank you first. I'm really sorry that I don't have much wealth

If the dynamic reference system does not rotate, B also has the speed of fixed axis rotation under the reference system
If the moving reference system is rotating, the reference system has a rotation speed with B

As shown in the figure, a disc rotates at a uniform speed in the horizontal plane with an angular velocity of 4rad / s. a small object with a mass of 0.10kg can move with the disc at a position 0.1M away from the center of the disc, as shown in the figure. When the object moves at a uniform speed in a circle, the centripetal force is______ N. On the centripetal force of the object, a and B have different opinions: a believes that the centripetal force is equal to the static friction force of the disk against the object, pointing to the center of the circle; B believes that the object has a tendency to move forward, and the direction of static friction is opposite to the direction of relative motion, that is, backward, rather than perpendicular to the direction of motion, so the centripetal force cannot be static friction. Do you agree______ My opinion

The centripetal force of the object in circular motion is FN = Mr ω 2＝0.1 × zero point one × 16N=0.16N．
The object makes a circle and provides centripetal force by static friction, because the direction of centripetal force points to the center of the circle, and the direction of static friction points to the center of the circle. Agree with Party A
So the answer is: 0.16, a

One radius is R and the angular velocity is ω What is the moment of inertia of the disk? Is the moment of inertia about the central axis

This question is actually incomplete. It depends on what axis you rotate. If you rotate around an axis perpendicular to the disk passing through the center of the circle
Let the surface density of the disk be K
Take a ring with radius R and width DR on the disk, then the area of the ring is 2 Π RDR and the mass of the ring
dm=2K∏rdr
Definition of moment of inertia
J=∫r^2dm
J = ∫ 2 Π Kr ^ 3DR, the integral interval is [0.r]
J = Mr ^ 2 / 2

As shown in the figure, the large disk with radius R is at angular velocity ω Rotation. If a person stands at the edge P of the plate and rotates with the plate, the force of the floor on the person is? To be explained in detail

From F = m ω^ 2R
The static friction force of the floor to people provides centripetal force, and the force is m ω^ The 2R direction points to the center of the circle
The support force of the floor to people is n = mg, and the direction is vertical and upward