Laws of Conservation of Energy and Momentum

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1. m.

2. .

J; J.

The initial values of kinetic and potential energy are respectively: The change in potential energy , where (taking s) is the change in the body's height over 1 s. The kinetic energy of the body increases by this same amount.

J.

At the end of the fourth second, the velocity of the body, thrown horizontally with speed from some height, will be composed of the horizontal velocity and the vertical velocity . The velocity of the body is found using the parallelogram rule (see fgigure). Hence, the kinetic energy of the body at the end of the fourth second is (see figure)

; ; .

The entire rope is accelerated by the force of gravity acting on the hanging part , where is the mass of the rope. According to Newton's second law, . Hence . The speed of the rope can be found from the law of conservation of energy. If a part of the rope of length slides off the table, its center of mass descends by , therefore . Hence and .

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See the solution to problem 5. Initially, the center of mass of the rope was at a distance from the peg, and at the moment the rope leaves the peg, its center of mass will be at a distance from the peg.

m.

The decrease in the mechanical energy of the skater is equal to the work done against friction: Here is the height to which the skater ascends; on the other hand, where is the path traveled by the skater along the inclined plane. Substituting expression (2) into equation (1), we find that .

J.

The work done by resistance forces is equal to the change in the mechanical energy of the body:

J.

See problem 8. The work done against resistance forces is positive.

N.

The work done by resistance forces . On the other hand, ( is the average resistance force, which is directed opposite to the displacement, is the length of the slope).

J.

From the law of conservation of energy, it follows that where is the amount of heat released during the impact.

; ; oscillatory.

From the equilibrium condition of the load , we find the spring stiffness . Now writing the law of conservation of energy and substituting the expression for into the equation, we find The speed of the load is maximum when the acceleration is zero and . Since , this occurs at . The energy conservation at this point is Substituting , we get so . The load will oscillate around the equilibrium position with amplitude (from to ).

N; s.

The work done against the resistance force of the soil is equal to the change in potential energy of the pile: . Hence N. The impact time is determined from the relation , where is the speed of the pile at the beginning of the impact (we assume the motion of the pile is uniformly decelerated, so ). In accordance with the law of conservation of energy . Therefore s.

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At the end of the path, the sled stops, and, consequently, all the initial potential energy is spent on work against friction forces on the inclined and horizontal sections of the path: where is the length of the slope; is the angle of inclination of the hill. In this case . Therefore, from equation (1) we get , whence .

a) N·s;

b) J.

The impulse of the force acting on the ball during the impact with the plate, in accordance with Newton's second law, is equal to: where and are the speeds of the ball respectively before and after the impact. From the law of conservation of energy, we find: and , therefore . According to Newton's third law, the impulse of the force that acted on the plate during the impact is numerically equal to . The amount of heat released during the impact of the ball on the plate is equal to the difference in the ball's energies before and after the impact:

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It is more convenient to solve the problem by considering the collision in a coordinate system associated with the moving plate. Since the speed of the plate changes negligibly due to the impact, this coordinate system can be considered inertial. In it, the ball before the impact with the plate has speed , where is the speed of the ball in the stationary coordinate system associated with the Earth. The impact of the ball on the plate is perfectly elastic, therefore, after the impact, the ball in the moving system will have speed . In the coordinate system associated with the Earth, the speed of the ball will be (the minus sign means that the ball will have a velocity directed upwards). Therefore, after the impact, the ball will bounce to a height

The potential energy of the ball-air-Earth system decreased, because as the ball rises, the volume occupied by the ball is replaced by air, which has a greater mass than the ball.

m.

During the motion of the puck, its kinetic energy is spent on doing work against friction:

In the first case, the speed of the body is greater than in the second.

Let denote the minimum speed with which the person must jump to reach the shore. During the jump, the person imparts a velocity to the boat (or steamship), equal, based on the law of conservation of momentum, to where is the mass of the person and is the mass of the boat or steamship.

By jumping, the person performs work , equal to the kinetic energy acquired by the person and the boat (or steamship):

This work is smaller, the smaller the ratio of the masses of the person and the boat (steamship). The mass of the steamship is many times greater than the mass of the person (), therefore, jumping from it, the person performs less work than in the case when they jump from the boat ().

m.

Based on the laws of conservation of momentum and energy, we can write the equations: s = \left( \dfrac{m}{M} \right)^2 \dfrac{v^2}{2kg}.A = \dfrac{mv_1^2}{2} + \dfrac{Mv_2^2}{2}, A = \dfrac{mv_1^2}{2} + \dfrac{m^2v_1^2}{2M} = \dfrac{mv_1^2}{2} \left( 1 + \dfrac{m}{M} \right). Mgh = Mv^2/2, n = 1000l = 1 \, \text{m}v\alpha = 10^\circv = (1 + n) \sqrt{4gl} \sin(\alpha/2) \approx 550m_1 = 10 \, \text{g}v_1 = 600 \, \text{m/s}m_2 = 0.5 \, \text{kg}s = 10 \, \text{cm}F_cs_1F_c = \dfrac{m_1 m_2 v_1^2}{2(m_1 + m_2)s} \approx 1.8 \times 10^4s_1 = s \dfrac{m_1 + m_2}{m_2} \approx 10.2u_x = 0.5nv_xv_{0x} = 19.5n = 19v_x = 0.5v_{0x} - u_x-(v_{0x} - u_x) - u_x = v_{0x} - 2u_xn(1)(2)sd = 4F = 30\rho = 1000s \approx 4.945^{\circ}s = v^2/gvmv^2/2Fllm = \rho SlS = \pi d^2/4\rhoFl = \rho Sl v^2/2v = \sqrt{2Fl/(\rho Sl)} = \sqrt{2F/(\rho S)} = \sqrt{8F/(\pi d^2 \rho)}s = 8F / (\rho g \pi d^2)DuFd\rhou = \dfrac{d^2}{D^2} \sqrt{\dfrac{8F}{\pi \rho (D^4 - d^4)}}u\Delta t\Delta m = \rho S u \Delta tS = \pi D^2/4Fu \Delta t = \dfrac{\Delta m v^2}{2} - \dfrac{\Delta m u^2}{2}vuv/u = S/s = D^2/d^2s = \pi d^2/4m_1m_2l_1/l_2 = m_2/m_1l_1l_2$ are the lengths of the arcs of the ring from the starting point to the point of the 11th collision.</p></details><details><summary>Hint</summary><p>The initial momenta of the beads are equalm_1v_{01} = m_2v_{02},mv_0 = -\dfrac{1}{2}mv_0 + Mv.v = \dfrac{3}{2} \dfrac{m}{M} v_0.
\begin{aligned}
E&=\frac{m,v_0^2}{2}-\frac{m,(v_0/2)^2}{2}-\frac{M,v^2}{2}\
&=\frac{3,m,v_0^2,(M-3m)}{8,M}.
\end{aligned}
m_1v_1 = m_2v_2 = m_3v_3. \tag{1 \& 2} m_1v_1^2 + m_2v_2^2 + m_3v_3^2 = 2E_0. \tag{3} m\vec{v}_1 = m\vec{v}_1' + m\vec{v}_2', \tag{1} \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_1'^2 + \dfrac{1}{2}mv_2'^2, \tag{2} \vec{v}_1 = \vec{v}_1' + \vec{v}_2', v_1^2 = v_1'^2 + v_2'^2.
m_B v = m_A v_{Ax}

m_B \cdot 0.5v = m_A v_{Ay}

v_A = \sqrt{v_{Ax}^2 + v_{Ay}^2} \approx 1.1v,\frac{m_B}{m_A}

m_B v^2 = m_B \cdot 0.25v^2 + 1.25v^2 \cdot \frac{m_B^2}{m_A^2} \cdot m_A

\frac{m_B}{m_A} = \frac{3}{5}
\quad\Rightarrow\quad v_A = 0.66v.

mv = mv_1 \cos\alpha + \frac{m}{2}v_2 \cos\beta
\quad \Rightarrow \quad
2v - 2v_1 \cos\alpha = v_2 \cos\beta \tag{1}

0 = mv_1 \sin\alpha - \frac{m}{2}v_2 \sin\beta
\quad \Rightarrow \quad
2v_1 \sin\alpha = v_2 \sin\beta \tag{2}

mv^2 = mv_1^2 + \frac{m}{2}v_2^2
\quad \Rightarrow \quad
2v^2 - 2v_1^2 = v_2^2 \tag{3}

3v_1^2 - 2\sqrt{3}vv_1 + v^2 = 0

v_1 = \frac{\sqrt{3}v}{3}

v_2 = \frac{2\sqrt{3}v}{3}

\beta = 30^\circ
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