Laws of Conservation of Energy and Momentum

$\beta =\arccos (2\cos \alpha -1)$.

1. $h=v_0^2/(4g)\approx 6.5$ m.

2. $v_0=\sqrt{2gh}$.

$T=32.2$ J; $\mathrm{\Pi }=39.4$ J.

The initial values of kinetic and potential energy are respectively: $$T=m{v}_0^2/2\$,\mathrm{\Pi }=mgH.$$ The change in potential energy $\mathrm{\Delta }\mathrm{\Pi }=mg\mathrm{\Delta }h$, where $\mathrm{\Delta }h=g{t}^2/2$ (taking $t=1$ s) is the change in the body's height over 1 s. The kinetic energy of the body increases by this same amount.

$T=1000$ J.

At the end of the fourth second, the velocity of the body, thrown horizontally with speed $v_0$ from some height, will be composed of the horizontal velocity $v_0$ and the vertical velocity $v_y=gt$. The velocity of the body $\overrightarrow{v}$ is found using the parallelogram rule (see fgigure). Hence, the kinetic energy of the body at the end of the fourth second is $$T=\frac{m{v}^2}{2}=\frac{m}{2}[v_0^2+(gt{)}^2].$$ (see figure)

$a=xg/L$; $v=x\sqrt{g/L}$; $v_{final}=\sqrt{gL}$.

The entire rope is accelerated by the force of gravity acting on the hanging part $F_g=mxg/L$, where $m$ is the mass of the rope. According to Newton's second law, $mxg/L=ma$. Hence $a=xg/L$. The speed of the rope can be found from the law of conservation of energy. If a part of the rope of length $x$ slides off the table, its center of mass descends by $h=x/2$, therefore $mg{x}^2/(2L)=m{v}^2/2$. Hence $v=x\sqrt{g/L}$ and $v_{final}=\sqrt{gL}$.

$v=\sqrt{gL/2}$.

See the solution to problem 5. Initially, the center of mass of the rope was at a distance $L/4$ from the peg, and at the moment the rope leaves the peg, its center of mass will be at a distance $L/2$ from the peg.

$H\approx 2$ m.

The decrease in the mechanical energy of the skater is equal to the work done against friction: \dfrac{mv^2}{2} - mgH = A_{fr} \tag{1} Here $H$ is the height to which the skater ascends; on the other hand, \begin{aligned} A_{fr} &= kmgl \cos \alpha \\ &= k \dfrac{H}{\sin \alpha} mg \cos \alpha \\ &= k \dfrac{mgH}{s} \end{aligned}\tag{2} where $l$ is the path traveled by the skater along the inclined plane. Substituting expression (2) into equation (1), we find that $$H=\frac{v^2}{2g(1+k/s)}$$.

$A\approx -80.2$ J.

The work done by resistance forces is equal to the change in the mechanical energy of the body: $$A=E_2-E_1=\frac{m{v}_2^2}{2}-(mgh+\frac{m{v}_1^2}{2}).$$

$A\approx 3.5$ J.

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

$F_{res}\approx 1570$ N.

The work done by resistance forces $A={\displaystyle \frac{m{v}_2^2}{2}}-({\displaystyle \frac{m{v}_1^2}{2}}+mgH)$. On the other hand, $A=-F_{res}l$ ($F_{res}$ is the average resistance force, which is directed opposite to the displacement, $l$ is the length of the slope).

$Q\approx 9.8$ J.

From the law of conservation of energy, it follows that $$mgh+\frac{m{v}_0^2}{2}=kmgs+Q,$$ where $Q$ is the amount of heat released during the impact.

$x_{\text{max}}=2x_0$; $v_{\text{max}}=\sqrt{g{x}_0}$; oscillatory.

From the equilibrium condition of the load $mg=k{x}_0$, we find the spring stiffness $k=mg/x_0$. Now writing the law of conservation of energy $$k{x}^2/2=mgx$$ and substituting the expression for $k$ into the equation, we find $$x=2x_0.$$ The speed of the load is maximum when the acceleration is zero and $mg=kx$. Since $mg=k{x}_0$, this occurs at $x=x_0$. The energy conservation at this point is $$\frac{m{v}_{\text{max}}^2}{2}+\frac{k{x}_0^2}{2}=mg{x}_0.$$ Substituting $k=mg/x_0$, we get $$\frac{m{v}_{\text{max}}^2}{2}+\frac{mg{x}_0}{2}=mg{x}_0,$$ so $v_{\text{max}}=\sqrt{g{x}_0}$. The load will oscillate around the equilibrium position $x_0$ with amplitude $x_0$ (from $x=0$ to $x=2x_0$).

$F_{res}=Mg(1+h/s)\approx 3\times {10}^5$ N; $\tau =2s/\sqrt{2gh}\approx 8\times {10}^{-3}$ s.

The work done against the resistance force of the soil is equal to the change in potential energy of the pile: $F_{res}s=Mg(h+s)$. Hence $F_{res}=Mg(1+h/s)=Mg{\displaystyle \frac{h+s}{s}}\approx 3\times {10}^5$ N. The impact time is determined from the relation $\tau =s/v_{avg}=2s/v_0$, where $v_0$ is the speed of the pile at the beginning of the impact (we assume the motion of the pile is uniformly decelerated, so $v_{avg}=v_0/2$). In accordance with the law of conservation of energy $v_0=\sqrt{2gh}$. Therefore $\tau =2s/\sqrt{2gh}\approx 8\times {10}^{-3}$ s.

$k=0.05$.

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: mgh = kmgs \cos \alpha + kmgl, \tag{1} where $s$ is the length of the slope; $\alpha$ is the angle of inclination of the hill. In this case $s\cos \alpha =b$. Therefore, from equation (1) we get $h=k(b+l)$, whence $k=h/(b+l)$.

a) $p=0.17$ N·s;

b) $Q=3.7\times {10}^{-2}$ 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: $$p=|\mathrm{\Delta }(m\overrightarrow{v})|=m(v_1+v_2),$$ where $v_1$ and $v_2$ are the speeds of the ball respectively before and after the impact. From the law of conservation of energy, we find: $v_1=\sqrt{2g{h}_1}$ and $v_2=\sqrt{2g{h}_2}$, therefore $p=m\sqrt{2g}(\sqrt{h_1}+\sqrt{h_2})$. According to Newton's third law, the impulse of the force that acted on the plate during the impact is numerically equal to $p$. 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: $$Q=\frac{m{v}_1^2}{2}-\frac{m{v}_2^2}{2}=mg(h_1-h_2).$$

$h=(\sqrt{2gl}+2u{)}^2/(2g)$.

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 v' = v + u, where $v=\sqrt{2gl}$ 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 v_1' = -v' = -(v + u). In the coordinate system associated with the Earth, the speed of the ball will be $v_1=-(v+u)-u=-(v+2u)$ (the minus sign means that the ball will have a velocity directed upwards). Therefore, after the impact, the ball will bounce to a height $$h=\frac{v_1^2}{2g}=\frac{(v+2u{)}^2}{2g}=\frac{(\sqrt{2gl}+2u{)}^2}{2g}.$$

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.

$l\approx 2.7$ m.

During the motion of the puck, its kinetic energy is spent on doing work against friction: $$m{v}_0^2/2=kmg(s+l).$$

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

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

By jumping, the person performs work $A$, equal to the kinetic energy acquired by the person and the boat (or steamship):
$$A=\frac{m{v}^2}{2}+\frac{M{u}^2}{2}=\frac{m{v}^2}{2}(1+\frac{m}{M}).$$
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 ($m/M\ll 1$), therefore, jumping from it, the person performs less work than in the case when they jump from the boat ($m/M\approx 1$).

$s={\left({\displaystyle \frac{m}{M}}\right)}^2{\displaystyle \frac{v^2}{2kg}}\approx 0.29$ m.

Based on the laws of conservation of momentum and energy, we can write the equations: Mv' = mv\quad\text{ and }\quad M\frac{(v')^2}{2} = kMgs$. Excluding $v'$ from these equations, we finds = \left( \dfrac{m}{M} \right)^2 \dfrac{v^2}{2kg}.A = \dfrac{mv_1^2}{2} + \dfrac{Mv_2^2}{2}, $$where\$v_2\$isthespeedofthecart.Bythelawofconservationofmomentum,\$m{v}_1=M{v}_2\$,whence\$v_2=m{v}_1/M\$;substituting\$v_2\$intotheexpressionforwork,weget$$ 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$,\qquad v^2 = 2gh.$ At the moment of firing, only the pressure forces of the powder gases act on the rifle and the bullet. These are internal forces. Therefore, the rifle-bullet system can be considered closed and the law of conservation of momentum can be applied to it. Before the shot, the momentum of the system was zero, therefore, after the shot, the total momentum of the system must also be zero: $M\vec{v} + m\vec{v}_1 = 0.$ Hence $v_1 = -\dfrac{Mv}{m} = -\dfrac{M}{m} \sqrt{2gh}.The '$-$' sign means that the bullet's velocity is directed opposite to the rifle's recoil velocity.

$\alpha \approx 15^{\circ}$.

The block, having received velocity $u$ upon impact by the bullet, rises in accordance with the law of conservation of mechanical energy to a height $h = u^2/(2g)$. On the other hand, $h = l(1 - \cos \alpha) = 2l \sin^2 (\alpha/2)$. Therefore $\sin^2 (\alpha/2) = u^2/(4gl)$; $u$ can be determined from the law of conservation of momentum:m_1v = (m_1 + m_2)u$, $u = \dfrac{m_1}{m_1 + m_2} v.$ Substituting the expression for $u$ into the equation for the sine of the angle, we find $ \sin \dfrac{\alpha}{2} = \dfrac{m_1v}{m_1 + m_2} \dfrac{1}{\sqrt{4gl}}.

$v = (1 + n) \sqrt{4gl} \sin(\alpha/2) \approx 550$ m/s.

See the solution to problem 24.

$F_c = \dfrac{m_1 m_2 v_1^2}{2(m_1 + m_2)s} \approx 1.8 \times 10^4$ H; $s_1 = s \dfrac{m_1 + m_2}{m_2} \approx 10.2$ cm.

Let's write the laws of conservation of energy and momentum for the bullet-block system: \dfrac{m_1 v_1^2}{2} = \dfrac{(m_1 + m_2)v^2}{2} + F_c s; \tag{1}$ $ m_1 v_1 = (m_1 + m_2)v, \tag{2} $$where\$v\$isthevelocityacquiredbytheblock.Solvingthesystemofequations(1)and(2),wefind\$F_c\$.Whenthebullethitsafixedblock,thelawofconservationofenergyiswrittendifferently:\$m_1{v}_1^2/2=F_c{s}_1\$.Substitutingtheexpressionfor\$F_c\$here,wefind$$s_1 = s \dfrac{m_1 + m_2}{m_2}.Explain what significance the length of the thread suspension has.

$v = \dfrac{M+m}{m \sin \alpha} \sqrt{2gh} \approx 219$ m/s.

Since the ball at the moment of impact can only move in the horizontal direction (see figure), applying the law of conservation of momentum to the bullet-ball system, we must consider only the horizontal component of the momentum: (M+m)v_1 = mv \sin \alpha. \tag{1} $$Bythelawofconservationofmechanicalenergy\$(M+m)v_1^2/2=(M+m)gh\$,andconsequently,\$v_1^2=2gh\$.Substitutingthisvalueintoequation(1),wegetthebulletspeed$$ v = \dfrac{M+m}{m \sin \alpha} \sqrt{2gh}. (m_1 + m_2) g h_{cm} = m_1gh - m_2gh, $$where\$m_1\$and\$m_2\$arethemassesoftheweights;\$h_{cm}\$istheheighttowhichthecenterofmassofthesystemdescends(rises);\$h\$isthedistancebywhicheachoftheweightshasrisen(descended).But\$h_{cm}=a_{cm}{t}^2/2\$;\$h=a{t}^2/2\$,where\$a\$istheaccelerationoftheweights:$$ a = \dfrac{m_1 - m_2}{m_1 + m_2} g. $ Therefore $ a_{cm} = \dfrac{m_1 - m_2}{m_1 + m_2} a = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2 g. $

2nd method. According to the law of motion of the center of mass, one can write
$ (m_1 + m_2) a_{cm} = m_1g + m_2g - 2T, \tag{1} $$where\$a_{cm}\$istheaccelerationofthecenterofmass;\$T\$isthetensionforceofthethread.Fromtheequationsofmotionoftheweights,itiseasytofindthat$$ T = \dfrac{2m_1m_2}{m_1 + m_2} g. \tag{2} $
Substitute expression (2) into equation (1):
$ a_{cm} = \left( \dfrac{m_1 - m_2}{m_1 + m_2} \right)^2 g.

$\eta = M/(M+m) \approx 93\%$.

By definition, efficiency is \eta = A_{useful} / A_{expended}. \tag{1} $$Inthiscase,theusefulworkwillbethework\$A_{useful}\$onthedeformationoftheblank,andtheexpendedwork\$A_{expended}\$isthekineticenergythatthehammerpossessedbeforetheimpact:\$A_{expended}=T_{initial}=m{v}^2/2\$.Thedeformationoftheblankendsatthemomentwhenthespeedofthehammerequalsthespeedoftheanvil,i.e.,atthemomentwhentheimpactitselfends.Thehammer-blank-anvilsystemcanbeconsideredclosedwithgoodaccuracy,sincetheanvilismountedonafoundationthroughshockabsorbers,dampingthevibrationsoftheanvil,andduringtheshortimpacttime,alargeadditionalforcecannotdevelopinthemduetothesmalldisplacementoftheanvilduringthistime.Thus,themomentumthatthehammerinitiallypossessedisconserved,andasaresultoftheinelasticimpact,ittransferstothesystemasawhole:$$ mv = (m + M)u, \tag{2} where $u$ is the joint speed of the system at the moment the impact ends. The 'residual' kinetic energy $T_k$ associated with this motion is the unused (lost) part of the initial energy $T_{initial}$. Then the useful work $A_{useful} = T_{initial} - T_k$, and the efficiency is \eta = \dfrac{A_{useful}}{A_{expended}} = \dfrac{T_{initial} - T_k}{T_{initial}} = 1 - \dfrac{T_k}{T_{initial}}. $$Usingformula(2),weget\$T_k=(M+m)u^2/2=m^2{v}^2/[2(M+m)]\$,andfinally$$ \eta = 1 - \dfrac{m^2v^2/[2(M+m)]}{mv^2/2} = 1 - \dfrac{m}{M+m} = \dfrac{M}{M+m}. Mv = (M+m)u; \tag{1} $ $ \dfrac{(M+m)u^2}{2} - \dfrac{Mv^2}{2} = -kmgS, \tag{2} $ where $S$ is the distance traveled by the block on the platform. The second expression equates the change in kinetic energy of the system (difference between final and initial values) to the work of the friction forces, which serves as a measure of the conversion of kinetic energy into internal energy (heat). Excluding $u$ from (1) and (2), we find $ S = \dfrac{M^2 v^2}{2kg(M+m)^2}. (Note: The original text has a different expression for S. Re-deriving: From (1), $u = Mv/(M+m)$. Substitute into (2): $\dfrac{(M+m)}{2} \dfrac{M^2v^2}{(M+m)^2} - \dfrac{Mv^2}{2} = -kmgS$. $\dfrac{M^2v^2}{2(M+m)} - \dfrac{Mv^2(M+m)}{2(M+m)} = -kmgS$. $\dfrac{Mv^2(M - (M+m))}{2(M+m)} = -kmgS$. $\dfrac{Mv^2(-m)}{2(M+m)} = -kmgS$. $S = \dfrac{mMv^2}{2kg(M+m)}$. The condition of the problem is met if $S \le l$. Therefore, $l_{min} = S = \dfrac{mMv^2}{2kg(M+m)}$. The amount of heat released $Q = kmgS$ is equal to the decrease in kinetic energy of the system: $Q = \dfrac{Mv^2}{2} - \dfrac{(M+m)u^2}{2} = \dfrac{Mv^2}{2} - \dfrac{M^2v^2}{2(M+m)} = \dfrac{Mv^2(M+m) - M^2v^2}{2(M+m)} = \dfrac{Mmv^2}{2(M+m)}$. Solve the problem another way, determining the accelerations of the block and the platform.

$s = \dfrac{mMv^2}{2kg(m+M)^2} \approx 0.25$ m; $Q = \dfrac{mMv^2}{2(m+M)} \approx 50$ J.

From the law of conservation of momentum $mv = (m+M)u$, we find $u = mv/(M+m)$ - the speed of the cart after the body stops. The path traveled by the cart is $s = u^2/(2a)$, where $a = kmg/M$ is the acceleration of the cart. The amount of heat released is determined from the law of conservation of energy: Q = \dfrac{mv^2}{2} - \dfrac{(m+M)u^2}{2} = \dfrac{mMv^2}{2(m+M)}. (m_1 + m_2)gh = (m_1 + m_2) \dfrac{u^2}{2}. \tag{1} $ We find the speed $u$ using the law of conservation of momentum: $ m_1v_1 = (m_1 + m_2)u. \tag{2} $$From(1),\$h=u^2/(2g)\$.Substituting\$u\$from(2):\$h=\frac{1}{2g}{\left(\frac{m_1{v}_1}{m_1+m_2}\right)}^2\$.Theamountofheat\$Q\$releasedduringtheimpactis$$ Q = \dfrac{m_1v_1^2}{2} - \dfrac{(m_1+m_2)u^2}{2} = \dfrac{m_1v_1^2}{2} - (m_1+m_2)gh = \dfrac{m_1v_1^2}{2} \left( 1 - \dfrac{m_1}{m_1+m_2} \right) = \dfrac{m_1m_2v_1^2}{2(m_1+m_2)}. v_{nx} = v_{0x} - 2nu_x. m_1 g\,h \;=\; m_2 g\,h_2 \;+\; m_1 g\,h_1, \tag{1} $ $ m_1\sqrt{2gh}\;=\;m_2\sqrt{2gh_2}\;+\;m_1\sqrt{2gh_1}. \tag{2} $ Transforming $(1)$ and $(2)$ as in Problem 28 and solving them jointly gives $ \sqrt{h_2} \;=\;\sqrt{h_1} \;+\;\sqrt{h}. \tag{3} $ Substitute $(3)$ into $(2)$ to find $ h_1 \;=\; h\,\Bigl(\frac{m_1 - m_2}{m_1 + m_2}\Bigr)^{2}. \tag{4} $ Finally, using $(1)$ together with $(4)$ yields $ h_2 \;=\; 4\,h\,\Bigl(\frac{m_1}{m_1 + m_2}\Bigr)^{2}. $

b) (See Problem 34.)
$
H ;=; h,\Bigl(\frac{m_1}{m_1 + m_2}\Bigr)^{2}.

$n = 5/3$.

The speed of the helium atom after the collision $v_1' = v_1 \dfrac{m_1 - m_2}{m_1 + m_2}$ (see problem 5.28), where $m_1$ is the mass of the helium atom; $m_2$ is the mass of the hydrogen atom; $v_1$ is the speed of the helium atom before the collision; $n = v_1 / |v_1'| = \dfrac{m_1+m_2}{|m_1-m_2|} = \dfrac{m_1/m_2 + 1}{|m_1/m_2 - 1|}$. (Note: $m_1 \approx 4u, m_2 \approx 1u$, so $n = (4+1)/|4-1| = 5/3$).

$E_1 = \dfrac{4\alpha}{(1+\alpha)^2} E_0$ (see figure).

From the laws of conservation of energy and momentum, it follows that the energy transferred to the initially stationary ball is E_1 = \dfrac{4\alpha}{(1+\alpha)^2} E_0, \dfrac{m_1v_1^2}{2} + \dfrac{m_2v_2^2}{2} = \dfrac{(m_1+m_2)u^2}{2} + \Pi_{\text{max}}; \tag{1} $ $ m_1v_1 + m_2v_2 = (m_1+m_2)u. \tag{2} $ From equation (2), $u = \frac{m_1v_1 + m_2v_2}{m_1+m_2}.$ Substituting this expression into equation (1), we obtain $\Pi_{\text{max}} = \dfrac{1}{2} \dfrac{m_1m_2}{m_1+m_2} (v_1 - v_2)^2. mg(h+h_1) = F_{\text{buoy}} h_1. \tag{1} On the right side of equality (1) is the work of the buoyant force $F_{\text{buoy}}$ during the submersion of the body. According to Archimedes' principle F_{\text{buoy}} = V_{body} g \rho_2 = mg \rho_2 / \rho_1, \tag{2} where $m = V_{body} \rho_1$ is the mass of the body. Substituting expression (2) into equation (1), we get $h_1 = h \rho_1 / (\rho_2 - \rho_1)$. The time $t$ can be found from the equality of the change in the body's momentum upon immersion to the impulse of the force: (F_{\text{buoy}} - mg)t = m\sqrt{2gh}. \tag{3} From relations (3) and (2) we find $t = \dfrac{m\sqrt{2gh}}{mg(\rho_2/\rho_1 - 1)} = \sqrt{\dfrac{2h}{g}} \dfrac{\rho_1}{\rho_2 - \rho_1}$. The acceleration (according to Newton's second law) is equal to $a = \dfrac{F_{\text{buoy}} - mg}{m} = g \dfrac{\rho_2 - \rho_1}{\rho_1}$.