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The board with mass m is fixed on a smooth horizontal plane. The bullet with mass m flies horizontally through the block at the speed V0 and the speed is V0 / 2. Now the block is not fixed and can slide on the smooth horizontal plane. The same bullet hits the block at the initial speed V0 in the horizontal direction. If M < 3M, then the bullet: A. Can shoot through a block of wood B. Can not shoot through the block, stay in the block to move together C. It just penetrates the block, but it stays at the edge of the block and moves together
Because the bullet through the board will be subject to the friction of the board on the bullet, and consume part of the energy, so as long as the energy is greater than this part of the energy, it can penetrate the board
At the beginning, the board is fixed. At this time, the horizontal direction has a strong effect, so momentum is not conserved, so according to the kinetic energy theorem
For bullets: MV0 & sup2 / 2 (V0 / 2) & sup2 / 2 = EF, where EF is the work done by friction, so EF = 3mv0 & sup2 / 8
If the bullet will stay in the board, it will move together with the board and have the same final velocity. The momentum in the horizontal direction of the system composed of the board and the bullet is conserved (where the friction is the internal force of the system), so it is possible
MV0 = (M + m) V (V is the common velocity of the board and the bullet) ②
Then we use the theorem of kinetic energy once more: MV0 & sup2 / 2 - (M + m) V & sup2 / 2 = EF 3
You can compare the above three equations. It's better to make m = 3M
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