1, the mechanical energy is conserved, the gravitational potential energy of the ball is converted into the kinetic energy of the ball and the car, and the momentum in the horizontal direction is conserved. Let the final ball speed v0 and the vehicle speed v 1 have mgL=m*v0 square /2+m * v/kloc-0 square/2, and 0 = m * v0+m * V6. Solve v 1 in formula 2 and substitute it into formula 1, and get: v0= radical sign [2Mgl/(M+m)].
2. The rope began to loosen, indicating that it has turned to a certain extent. The centripetal acceleration of the ball just doesn't need the tension of the rope to provide. In this problem, the component of gravity can provide centripetal acceleration. Let the angle x satisfy the condition (0
-mg*cosx=mv 1 square/l.
From the formula 1, it is easy to get m*v 1 square =m*v0 square -2mg*(L-L*cos x)=2mgL+2mgL*cosx. Substituting into Type 2, the solution is:
Cosx=-2/3 is approximately equal to -0.66667, and X = 13 1 49' from the table lookup.
3. This problem, let's talk about the work done by the wood block on the spring, 1 integration method. The integrand function is: 100x, the lower integral limit is 0, and the upper integral limit is 0. 1. The work done in solving this problem is 0.5J In high school, we tried to solve this kind of problem by averaging (professionally speaking, the integrand function is about 1).
First of all, there is an ideal situation in the title: "After the bullet is embedded, the spring is compressed 10cm", which means that the bullet hits the wood block for a short time, so that the wood block does not move during this time, so that the bullet and the wood block meet the conservation of momentum during the embedding process. Then it can be solved by the conservation of momentum. Let the initial velocity of the bullet be v0, the speed of the bullet just completely embedded in the wood block be v 1, the bullet mass be M, the wood block mass be M, the elastic potential energy of the spring be E, the dynamic friction coefficient be U, and the spring compression amount be L.
Conservation of energy, the kinetic energy of bullets and wood blocks is converted into elastic potential energy (0.5J was found before) and internal energy of springs, (m+M)*v 1 square /2=E+u(m+M)gl, and the solution is v1= 0.7149 m/.
Conservation of momentum: m*v0=(m+M)*v 1, and the solution is v0=32 1.7 (I take 10 from g, so maybe the answer is a little biased).
The third problem does not involve the situation that bullets can convert internal energy into depth, so it is not a difficult problem.
Did I explain it in detail? Please give me more.