Task. A puck of mass m slides with speed v 0 along a smooth horizontal surface of a table, hits a resting wedge of mass 2m, slides along it without friction or separation, and leaves the wedge (see figure). The wedge, which did not come off the table, acquires speed v 0 /4. Find the angle of inclination of the surface of the upper part of the wedge to the horizon. The lower part of the wedge has a smooth transition to the table surface. Neglect 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 Supporting abstract 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 lifting off the wedge. The kinetic energy of the wedge immediately after the puck comes off. (We neglect the change in the potential energy of the puck according to 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 onto this axis is preserved (see Supporting Note III, paragraph 5 vkotov.narod.ru/3. pdf) Projection of the puck's momentum before contact with the wedge. Projection of the puck's momentum immediately after liftoff from the wedge. Projection of the wedge impulse immediately after the puck comes off. 3. The modulus of the speed of the puck immediately after lifting off from the wedge v is related to the projections of this speed v x and v y: v 2 = v x 2 + v y 2 = (v 0 2 /4) + v y 2 Let’s substitute this into the formula for the law of conservation of energy (point 1) and after contractions we obtain: After contraction we obtain: v x = v 0 /2 4. In the moving reference frame X"O"Y" associated with the wedge, the speed of the puck immediately after liftoff v" will be directed at an angle to the horizontal. 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 velocities (see Supporting abstract I, paragraph 2 vkotov.narod.ru/1.pdf) The speed of the wedge immediately after lifting off the washer v k = v 0 /4 .
O X Y 5. Let us add the vectors v " and v k according to the triangle rule and in the same figure we show the decomposition of the vector v into the components v x and v y: The desired angle can be found from a triangle whose hypotenuse is v ", and the legs are parallel to the OX and OY axes. 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:
Physics problem - 2896
2017-04-16
A puck of mass $m$ slides with a speed of $v_(0)$ along a smooth horizontal surface of the table, hits a resting wedge of mass $2m$, slides along it without friction or separation, and leaves the wedge (Fig.). A wedge that does not leave the table will acquire a speed of $v_(0)/4$. Find the angle $\alpha$ 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. Neglect the change in the potential energy of the puck in the gravitational field as it moves along the wedge. The directions of all movements are parallel to the plane of the drawing. 
Solution:
The figure shows the moment the washer slides off the wedge. At this moment, let us denote the speed of the puck relative to the wedge by $\vec(v)_(rel)$, and the speed of the wedge itself by $\vec(u)$. It is obvious that the speed of the wedge is directed horizontally, and the relative speed of the puck makes an angle $\alpha$ with the horizon. Since the resulting force acting on the “washer plus wedge” system of bodies in the horizontal direction is zero, the horizontal component of the momentum of this system remains unchanged:
$mv_(0) = 2mu + m(v_(rel) \cos \alpha + u)$. (1)
Since $u = v_(0)/4$, then equation (1) will have the form
$v_(0) = 4 v_(rel) \cos \alpha$. (2)
According to the law of conservation of energy
$\frac(mv_(0)^(2))(2) = \frac(2mu^(2))(2) + \frac(mv_(w)^(2))(2)$. (3)
In this equation, $v_(w)$ is the speed of the puck at the moment of sliding relative to a fixed coordinate system. By the cosine theorem
$v_(w)^(2) = v_(rel)^(2) + u^(2) + 2 v_(rel) u \cos \alpha$.
After substituting this relation into (3) and taking into account the fact that $u = v_(0)/4$, we obtain
$13 v_(0)^(2) = 16 v_(rel)^(2) + 8 v_(rel) v_(0) \cos \alpha$. (4)
From the joint solution of (2) and (3) with respect to $\cos \alpha$ we obtain that
$\cos \alpha = \frac(1)( \sqrt(11))$.
Option No. 2819169
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