(1). The X direction is uniform linear motion, and the Y direction is uniformly accelerated linear motion, so the synthetic motion is quasi-planar motion. The motion trajectory is shown in the figure.

The pictures on the computer don't look good, so I'll make do with it. ...

(2) Substitute t= 1, 2 into X and Y expressions for calculation. When t= 1, the particle is located at (2, 1); when t=2, the particle is located at (4,2), and the displacement vector is (2,3).

(3) Derive x and y respectively, and the relationship between speed and time is vx=2 and vy=-2t, so when t= 1, the speed vector is (2, -2), its size is root number 8, and its direction is vector (1,-1). Similarly, when t=2, the velocity vector is (2, -4), its size is root number 20 and its direction is vector (1, -2).

(4) vx and vy are derived respectively, and the relationship between acceleration and time is ax=0 and ay=-2. So when t= 1, the acceleration is 2 and the direction points to the negative y axis. When t=2, the acceleration is 4 and the direction points to the negative Y axis.

2. Because the cylinder is infinitely long and the electric field intensity is axisymmetric, Gauss theorem is directly used.

(1). r < When R 1, because there is no charge in the Gaussian plane at this time, the electric field intensity is 0 everywhere.

(2) Take a Gaussian surface with a cylindrical length of L, because the electric field is axisymmetrical, there will be no vertical component. Just consider the outward component of radiation, and it is easy to get the expression of Gaussian theorem, 2πrlE=λl/ε, and calculate the expression of E (ε is the vacuum dielectric constant, and footer 0 is not easy to hit).

(3) R>R 2 believes that although there is charge in the Gaussian plane, the positive and negative cancel each other out, and the overall effect is the same as that without charge, so the electric field intensity is still zero everywhere.

3. Suppose the bullet hits the object A, analyze the bullet and the object A, and finally * * * and the velocity of the object A, and use the momentum conservation m*v0=(m+M)*v to get v=(m*v0)/(m+M). After analyzing the motion between the bullet and the object A as a whole, there is still momentum conservation, that is, when v'=v', the kinetic energy of the system is the smallest and the potential energy is the largest, that is, the energy of the spring is the largest. If we solve the equation (m+m) * v'+m * v' = m * v0, v'=v'', we can get v' = v' = (m * v0)/(m+. At this time, the kinetic energy of the computing system is ek = (1/2) * (m+2m) * (v' 2) = (1/2) * (m+2m) * {[(m * v0)/(m+2m)] 2. Analyzing the kinetic energy of the system when the bullet hits A, there are ek0 = (1/2) * (m+m) * (v2) = (1/2) * (m+m) * {[(m * v0)/(m+m)] 2. Then the difference between Ek0 and Ek is the elastic potential energy of the spring. According to the elastic potential energy formula of the spring, (1/2) * k * x 2 = ek0-ek, and x is the maximum compression degree of the spring. Get x = {[m * m * (v0) 2]/[k * (m+m).

Please forgive me for answering so late.