A washer of mass m. On the way of a trolley of mass m sliding on a smooth horizontal table with a speed

Task 1

pendulum thread length l\u003d 1 m, to which a load of mass m= 0.1 kg,
deflected by an angle a from the vertical position and released.
Thread tension force T at the moment the pendulum passes the equilibrium position is 2 N.
What is equal to the angle a?

Solution

Based on Newton's second law, acceleration
caused by the sum of the forces of gravity acting on the load and the tension of the thread,
when passing through the equilibrium position is equal to the centripetal acceleration:

According to the law of conservation of mechanical energy, for the weight of the pendulum
(the lower position of the load is chosen as the origin of the potential energy):

Answer in general view and in numerical form:

Task 2

A washer of mass m starts moving along the chute AB from point A from a state of rest.
Point A is located above point B at a height of H = 6 m.
In the process of moving along the chute, the mechanical energy of the washer decreases by ΔE = 2 J due to friction.
At point B, the puck flies out of the chute at an angle α = 15° to the horizon and falls to the ground at point D, which is on the same horizontal line as point B (see figure). BD = 4 m.
Find the mass of the puck m.


Ignore air resistance.

Solution

Task 3 (for independent solution)

A puck thrown along an inclined plane slides down it
moving up and then moving down.
The plot of the puck speed modulus versus time is given in the figure.
Find the angle of inclination of the plane to the horizon.

Task. A puck of mass m slides at a speed v 0 along a smooth horizontal surface of the table, hits a wedge of mass 2m at rest, slides along it without friction and separation, and leaves the wedge (see Fig.). The wedge, which did not come off the table, acquires the speed v 0 /4. Find the angle of inclination to the horizon of the surface of the upper part of the wedge. The lower part of the wedge has a smooth transition to the table surface. Ignore the change in the potential energy of the puck in the gravitational field as it moves along the wedge. Solution. 1. The work of non-conservative forces in the system is zero, therefore, the mechanical energy of the system is conserved (see Reference Note III, paragraph 10 vkotov.narod.ru/3.pdf) The kinetic energy of the washer before contact with the wedge (we take the potential energy of the washer in this position equal to zero) The kinetic energy of the puck immediately after leaving the wedge. The kinetic energy of the wedge immediately after the puck breaks off. (Change in the potential energy of the puck is neglected by the condition)


2. All external forces acting on the bodies of our system (gravity and the reaction force of the table) are perpendicular to the horizontal axis OX, therefore, the projection of the system's momentum on this axis is preserved (see Reference abstract III, p. 5 vkotov.narod.ru/3. pdf) Projection of the momentum of the washer before contact with the wedge. Projection of the momentum of the puck immediately after the release from the wedge. Projection of the momentum of the wedge immediately after the puck breaks off. 3. The speed modulus of the puck immediately after leaving the wedge v is related to the projections of this speed vx and vy: v 2 = vx 2 + vy 2 = (v 0 2 /4) + vy 2 Substitute this into the formula for the law of conservation of energy (point 1) and after contractions we get: After contraction we get: vx = v 0 /2 4. In the moving reference frame X"O"Y" associated with the wedge, the speed of the puck immediately after the separation v" will be directed at an angle to the horizon. The speed v" of the puck relative to the wedge is related to the speed v of the puck relative to the table according to the law of addition of speeds (see Reference Note I, p. 2 vkotov.narod.ru/1.pdf) The speed of the wedge immediately after the puck comes off v k \u003d v 0 / 4 .


О X Y 5. Let's perform the addition of vectors v "and v k according to the triangle rule and in the same figure we will show the decomposition of the vector v into components v x and v y: The required angle can be found from a triangle, the hypotenuse of which is v", and the legs are parallel to the axes OX and OY. The figure shows that tg v y /(v x v k) Substituting v x from point 2, v y from point 3 and v k from the problem data, we get the answer:

Option No. 2819169

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