Kinetics of particles: Energy and Momentum Methods

Central impact and its types

Central impact occurs when the centres of mass of the two colliding bodies lie on the line of impact (the common normal at the contact point) and their velocities are along that line.

• Direct central impact: both velocities are along the line of impact.
• Oblique central impact: one or both velocities are inclined to the line of impact; the components normal to it follow the impact laws, the tangential components stay unchanged (smooth bodies).

By the coefficient of restitution $e$:

Maximum deflection of the pan

Data: $m_A=40$ kg, $m_B=10$ kg, $h=1.5$ m, $k=25$ kN/m, $e=0$ (perfectly plastic), $g=9.81$.

Step 1: block just before impact

$$v_A=\sqrt{2gh}=\sqrt{2(9.81)(1.5)}=5.425\ \text{m/s}$$

Step 2: plastic impact (pan is initially at rest, momentum conserved):

$$40(5.425)=(40+10)v'\ \Rightarrow\ v'=4.340\ \text{m/s}$$

Step 3: initial spring compression (pan alone in equilibrium):

$$x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{25000}=0.003924\ \text{m}$$

Step 4: work-energy from just after impact to the lowest point (total compression $x_2$, mass $M=50$ kg):

$$\tfrac12 M v'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}$$

With $\tfrac12Mv'^2=470.88$ J this gives

$$12500x_2^{2}-490.5x_2-(470.88+0.1925-1.9247)=0$$ $$12500x_2^{2}-490.5x_2-469.15=0\ \Rightarrow\ x_2=0.2143\ \text{m}$$

The pan therefore moves down $x_2-x_1=0.2143-0.0039=0.2104$ m from its position at impact.

Answer: maximum deflection of the pan from its position at impact $\approx0.210$ m (total spring compression $\approx0.214$ m).

Data: $m_A=30$ kg, $m_B=10$ kg, $h=2$ m, $k=30$ kN/m, perfectly plastic impact ($e=0$), $g=9.81\ \text{m/s}^2$.

Step 1: block just before impact

$$v_A=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}$$

Step 2: plastic impact (pan initially at rest)

Momentum is conserved and both move together:

$$30(6.264)=(30+10)v'\ \Rightarrow\ v'=4.698\ \text{m/s}$$

Step 3: initial compression of the spring due to the pan

$$x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{30000}=0.00327\ \text{m}$$

Step 4: work-energy from just after impact to the lowest point

Let $x_2$ be the maximum total compression and $M=40$ kg. Kinetic energy just after impact: $\tfrac12Mv'^2=441.45$ J.

$$\tfrac12Mv'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}$$ $$15000x_2^{2}-392.40x_2-\left(441.45+0.1604-1.2831\right)=0$$ $$x_2=0.1849\ \text{m}$$

The pan moves down $x_2-x_1=0.1816$ m from its position at impact.

Answer: maximum deflection of the pan $\approx0.182$ m below its position at impact (total spring compression $\approx0.185$ m).

Data: $m_A=30$ kg, $m_B=10$ kg, $h=2$ m, $k=20$ kN/m, perfectly plastic impact ($e=0$), $g=9.81\ \text{m/s}^2$.

Step 1: block just before impact

$$v_A=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}$$

Step 2: plastic impact (pan initially at rest)

Momentum is conserved and both move together:

$$30(6.264)=(30+10)v'\ \Rightarrow\ v'=4.698\ \text{m/s}$$

Step 3: initial compression of the spring due to the pan

$$x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{20000}=0.00490\ \text{m}$$

Step 4: work-energy from just after impact to the lowest point

Let $x_2$ be the maximum total compression and $M=40$ kg. Kinetic energy just after impact: $\tfrac12Mv'^2=441.45$ J.

$$\tfrac12Mv'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}$$ $$10000x_2^{2}-392.40x_2-\left(441.45+0.2406-1.9247\right)=0$$ $$x_2=0.2302\ \text{m}$$

The pan moves down $x_2-x_1=0.2253$ m from its position at impact.

Answer: maximum deflection of the pan $\approx0.225$ m below its position at impact (total spring compression $\approx0.230$ m).

Variable system of particles

A variable system of particles is a system whose mass changes with time: particles continuously enter (are gained by) or leave (are lost from) the system, for example a rocket expelling gas, a jet of water striking a vane, or a conveyor belt receiving sand. The principles of impulse-momentum are applied to the particles that are in the system at time $t$ plus those that join it in $dt$.

Pool balls

Three equal balls (equal mass $m$). Ball A with $v_0=4.57$ m/s strikes B and C at rest. Reading of the figure (chosen to be consistent with the given $V_C=3.42$ m/s): use axes $x'$ along $v_0$ and $y'$ perpendicular to it. After impact C moves along $x'$ (the same line as $v_0$), A moves along $-y'$ (perpendicular to $v_0$) and B moves at $60^\circ$ to $x'$ on the side opposite to A.

        y'
        |   B (60 deg to x')
        |  /
        | /
  v0 -> +------------> C   (x')
        |
        A (along -y')

Momentum along $x'$:

$$m v_0=m V_C+m V_B\cos60^\circ\ \Rightarrow\ V_B=\frac{v_0-V_C}{\cos60^\circ}=\frac{4.57-3.42}{0.5}=2.30\ \text{m/s}$$

Momentum along $y'$:

$$0=m V_B\sin60^\circ-m V_A\ \Rightarrow\ V_A=2.30(0.866)=1.99\ \text{m/s}$$

Check with energy (perfectly elastic): initial $v_0^{2}=20.88$; final $V_A^{2}+V_B^{2}+V_C^{2}=3.97+5.29+11.70=20.96$. These agree within rounding of the given $V_C$, so the interpretation is consistent.

Answer: $V_A\approx1.99$ m/s, $V_B\approx2.30$ m/s.

Setup. All three balls have equal mass $m$, the table is frictionless and the impact is perfectly elastic, so linear momentum and kinetic energy are both conserved. Axes: $x$ horizontal, $y$ vertical. From the figure: $v_0=5$ m/s at $45^\circ$ above the horizontal, $V_A$ vertically upward, $V_B$ at $30^\circ$ below the horizontal and $V_C$ at $30^\circ$ above the horizontal (B and C on the same side).

Momentum in $x$:

$$m(5\cos45^\circ)=mV_B\cos30^\circ+mV_C\cos30^\circ$$ $$V_B+V_C=\frac{3.5355}{0.8660}=4.0825\qquad(1)$$

Momentum in $y$:

$$m(5\sin45^\circ)=mV_A-mV_B\sin30^\circ+mV_C\sin30^\circ$$ $$V_A=3.5355+0.5(V_B-V_C)\qquad(2)$$

Energy:

$$V_A^{2}+V_B^{2}+V_C^{2}=5^{2}=25\qquad(3)$$

Let $D=V_B-V_C$. From (1): $V_B^2+V_C^2=\dfrac{(4.0825)^2+D^2}{2}=8.333+0.5D^2$. From (2): $V_A^2=12.5+3.5355D+0.25D^2$. Substituting in (3):

$$0.75D^{2}+3.5355D-4.1667=0\ \Rightarrow\ D=0.9763$$

(the other root, $D=-5.69$, is rejected because it makes $V_C$ negative.)

Then

$$V_B=\frac{4.0825+0.9763}{2}=2.529\ \text{m/s},\qquad V_C=\frac{4.0825-0.9763}{2}=1.553\ \text{m/s}$$ $$V_A=3.5355+0.5(0.9763)=4.024\ \text{m/s}$$

Check: $4.024^2+2.529^2+1.553^2=16.19+6.40+2.41=25.0$.

Answer: $V_A=4.02$ m/s, $V_B=2.53$ m/s, $V_C=1.55$ m/s.

Conservation of energy in kinetics of particles

When a particle moves under conservative forces only (weight, elastic spring force), the sum of kinetic energy $T=\tfrac12mv^{2}$ and potential energy $V=V_g+V_e$ stays constant:

$$T_1+V_1=T_2+V_2,\qquad V_g=mgy,\quad V_e=\tfrac12kx^{2}$$

where $x$ is the spring deformation from its undeformed length. Forces that do no work (normal reaction of a smooth rod) can be ignored.

Collar problem

Assumed geometry (figure dimensions 150, 200, 250 mm): semicircular rod of radius $0.2$ m with A and C at the same level, B $0.2$ m below. The spring is fixed at D on the vertical axis through B, $0.15$ m below the line AC. Then the spring length is $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.2-0.15=0.05$ m. Undeformed length $0.10$ m; $k=800$ N/m; $m=5$ kg.

Spring deformations: $x_A=x_C=0.25-0.10=0.15$ m (stretched) and $x_B=0.05-0.10=-0.05$ m (compressed).

a) At B

Released from rest at A (take B as datum, $h_A=0.2$ m):

$$0+mgh_A+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+0+\tfrac12kx_B^{2}$$ $$5(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(5)v_B^{2}+\tfrac12(800)(0.05)^{2}$$ $$9.81+9.00=2.5v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=7.124,\ \ v_B=2.669\ \text{m/s}$$

b) At C

C is at the same height as A and the spring has the same length, so both potential energies are the same as at A:

$$\tfrac12mv_C^{2}=0\ \Rightarrow\ v_C=0$$

Answer: $v_B\approx2.67$ m/s and $v_C=0$ for the geometry assumed above. If the figure shows C at a different level or spring length, use the same energy equation with those values.

a) Principle of conservation of energy

The total mechanical energy of a system, kinetic plus potential, remains constant when only conservative forces (gravity, springs) do work:

$$T_1+V_1=T_2+V_2$$

Example: a pendulum bob released from rest at height $h$ above its lowest point. At the top it has only potential energy $mgh$; at the bottom only kinetic energy $\tfrac12mv^{2}$. Hence $v=\sqrt{2gh}$. During the swing the energy changes between the two forms but the total is constant (air friction neglected).

b) Collar problem

Assumed geometry (figure dimensions 150, 200, 250 mm): semicircular rod of radius $0.2$ m, A and C at the same level, B $0.2$ m below them. The spring is fixed at D on the vertical axis through B, $0.15$ m below line AC, so $L_A=L_C=\sqrt{0.2^{2}+0.15^{2}}=0.25$ m and $L_B=0.05$ m. Undeformed length $0.10$ m, $k=800$ N/m, $m=2$ kg.

Deformations: $x_A=x_C=0.15$ m, $x_B=-0.05$ m.

At B (datum at B, $h_A=0.2$ m):

$$2(9.81)(0.2)+\tfrac12(800)(0.15)^{2}=\tfrac12(2)v_B^{2}+\tfrac12(800)(0.05)^{2}$$ $$3.924+9.00=v_B^{2}+1.00\ \Rightarrow\ v_B^{2}=11.924,\ \ v_B=3.453\ \text{m/s}$$

At C: same height and same spring deformation as at A, so $\tfrac12mv_C^{2}=0$ and $v_C=0$.

Answer: $v_B\approx3.45$ m/s; $v_C=0$ for the assumed geometry (use the same energy equation with figure values if C lies at a different level).

Types of impact

Impact is a collision of two bodies in a very short time, producing large forces. The line through the contact point, normal to the surfaces, is the line of impact.

• Central impact: the centres of mass of both bodies lie on the line of impact. It is direct if velocities are along this line and oblique if they are inclined to it.
• Eccentric impact: the line of impact does not pass through the centre of mass of one or both bodies.
• By the coefficient of restitution $e$: perfectly elastic ($e=1$, no energy loss), perfectly plastic ($e=0$, bodies stick together) and partly elastic ($0<e<1$).

Collar on the vertical rod

Assumptions (figure not shown): the spring ($k=100$ N/m, free length 1 m) is attached to a wall point level with $s=0$ and 1.5 m horizontally from the rod. The collar moves down the rod. Only the weight and the spring do work.

 wall   1.5 m   rod
  o-------------A  s = 0
   \            |
    \  spring   |  s
     \          |
      \_________B

Spring stretches (using $\ell=\sqrt{1.5^{2}+s^{2}}$):

• at $s=0$: $\ell=1.5$ m, stretch $=0.5$ m
• at $s$: stretch $=\ell-1$

Work-energy from $s=0$ to the momentary stop ($v=0$):

$$\tfrac12(4)(3)^{2}+4(9.81)s-\tfrac12(100)\left(\sqrt{1.5^{2}+s^{2}}-1\right)^{2}+\tfrac12(100)(0.5)^{2}=0$$ $$18+39.24s+12.5=50\left(\sqrt{2.25+s^{2}}-1\right)^{2}$$

Solving numerically gives $s=1.955$ m. Check: $\ell=\sqrt{2.25+1.955^{2}}=2.464$ m, so $50(1.464)^{2}=107.2$ J and the left side is $30.5+76.7=107.2$ J. The check holds.

Answer: the collar stops momentarily after travelling $s\approx1.96$ m.

Impact and its types

Impact is the collision of two bodies during a very short time interval, with large forces acting. The common normal at the contact point is the line of impact.

• Central impact: centres of mass lie on the line of impact. It is direct (velocities along the line) or oblique (velocities inclined to it).
• Eccentric impact: the line of impact misses the centre of mass of a body.
• By $e$ (coefficient of restitution): perfectly elastic $e=1$, perfectly plastic $e=0$, partly elastic $0<e<1$.

Collar problem

Motion is in a horizontal plane, so weight does no work. The normal force from the smooth rod does no work. Only the spring does work, so

$$\tfrac12mv_1^{2}+\tfrac12k\delta_1^{2}=\tfrac12mv_2^{2}+\tfrac12k\delta_2^{2}$$

Assumptions (figure not legible):

• $k=400$ N/m (the value $400$ kN/m would make the collar speed unrealistically large).
• The spring's fixed end D is such that the spring lengths are $L_A=150$ mm, $L_B=\sqrt{150^{2}+200^{2}}=250$ mm and $L_C=100$ mm.
• Undeformed length $L_0=80$ mm, so $\delta_A=0.070$ m, $\delta_B=0.170$ m, $\delta_C=0.020$ m. $m=0.5$ kg.

(i) Speed at A so that the collar just reaches B

$$\tfrac12(0.5)v_A^{2}+\tfrac12(400)(0.070)^{2}=0+\tfrac12(400)(0.170)^{2}$$ $$0.25v_A^{2}=5.78-0.98=4.80\ \Rightarrow\ v_A=4.38\ \text{m/s}$$

(ii) Speed at C

From B (at rest) to C:

$$0+\tfrac12(400)(0.170)^{2}=\tfrac12(0.5)v_C^{2}+\tfrac12(400)(0.020)^{2}$$ $$5.78=0.25v_C^{2}+0.08\ \Rightarrow\ v_C=4.77\ \text{m/s}$$

Answer: (i) $v_A\approx4.38$ m/s; (ii) $v_C\approx4.77$ m/s (for the assumed geometry; the method is the energy balance above with the lengths read from the figure).

Impact of the two balls

Assumptions (figure garbled): the line of impact is the horizontal x-axis. Ball A moves to the right at $3$ m/s inclined $40^\circ$ above the line of impact; ball B moves to the left at $4.5$ m/s inclined $25^\circ$ below the line of impact. $m_A=6$ kg, $m_B=3$ kg, $e=0.67$.

Components (normal $n$ = along line of impact, tangential $t$):

Tangential components do not change (smooth balls): $v'_{At}=1.928$, $v'_{Bt}=-1.902$ m/s.

Normal direction: momentum and restitution.

$$6(2.298)+3(-4.078)=6v'_{An}+3v'_{Bn}\quad\Rightarrow\quad 2v'_{An}+v'_{Bn}=0.5185$$ $$v'_{Bn}-v'_{An}=e(v_{An}-v_{Bn})=0.67(2.298+4.078)=4.272$$

Solving: $v'_{An}=-1.251$ m/s and $v'_{Bn}=3.021$ m/s.

Final velocities:

$$v'_A=\sqrt{1.251^{2}+1.928^{2}}=2.30\ \text{m/s},\quad \text{at }57.0^\circ\text{ above the negative line of impact}$$ $$v'_B=\sqrt{3.021^{2}+1.902^{2}}=3.57\ \text{m/s},\quad \text{at }32.2^\circ\text{ below the positive line of impact}$$

Principle of work and energy

The work done by all forces acting on a particle as it moves from position 1 to 2 equals the change in its kinetic energy:

$$T_1+U_{1\to2}=T_2,\qquad T=\tfrac12mv^{2}$$

where $U_{1\to2}=\int\vec F\cdot d\vec r$. Work of the weight is $\pm mg\,\Delta h$, work of a spring is $\tfrac12k(x_1^2-x_2^2)$, friction does negative work. It is a scalar equation, needs no time, and connects speed with displacement. For conservative forces it becomes $T_1+V_1=T_2+V_2$.

Answer: $v'_A\approx2.30$ m/s, $v'_B\approx3.57$ m/s (for the assumed directions).

Assumptions (figure not shown): the x-axis is the line of impact. Ball A ($m_A=1.2$ kg, $20$ m/s) moves to the right, $30^\circ$ above the x-axis; ball B ($m_B=1.8$ kg, $25$ m/s) moves to the left, $45^\circ$ below the x-axis. The balls are smooth, $e=0.84$.

Components

The $y$ components are unchanged by the impact: $v'_{Ay}=10.00$ m/s, $v'_{By}=-17.68$ m/s.

Along the line of impact

Momentum:

$$1.2(17.32)+1.8(-17.68)=1.2v'_{Ax}+1.8v'_{Bx}\ \Rightarrow\ 1.2v'_{Ax}+1.8v'_{Bx}=-11.04$$

Restitution:

$$v'_{Bx}-v'_{Ax}=e(v_{Ax}-v_{Bx})=0.84(17.32+17.68)=29.40$$

Solving: $v'_{Ax}=-21.32$ m/s, $v'_{Bx}=8.08$ m/s.

Final velocities

$$v'_A=\sqrt{21.32^{2}+10^{2}}=23.55\ \text{m/s},\qquad \text{direction }25.1^\circ\text{ above the negative x-axis}$$ $$v'_B=\sqrt{8.08^{2}+17.68^{2}}=19.44\ \text{m/s},\qquad \text{direction }65.4^\circ\text{ below the positive x-axis}$$

Kinetic energy lost

$$\begin{aligned} T_1&=\tfrac12(1.2)(20)^{2}+\tfrac12(1.8)(25)^{2}=240+562.5=802.5\ \text{J}\\ T_2&=\tfrac12(1.2)(23.55)^{2}+\tfrac12(1.8)(19.44)^{2}=672.7\ \text{J}\\ \Delta T&=802.5-672.7=129.8\ \text{J} \end{aligned}$$

Answer: $v'_A=23.55$ m/s, $v'_B=19.44$ m/s; kinetic energy lost $\approx129.8$ J.

Assumptions (figure description is ambiguous): the line of impact is the horizontal x-axis. Ball A (30 kg) moves to the right at $30$ m/s. Ball B (50 kg) moves to the left and upward at $30$ m/s, $45^\circ$ to the line of impact. $e=0.9$, smooth balls.

Components before impact

$$v_{Ax}=30,\ v_{Ay}=0;\qquad v_{Bx}=-30\cos45^\circ=-21.21,\ v_{By}=30\sin45^\circ=21.21\ \text{m/s}$$

Tangential (y) components do not change: $v'_{Ay}=0$, $v'_{By}=21.21$ m/s.

Along the line of impact

Momentum:

$$30(30)+50(-21.21)=30v'_{Ax}+50v'_{Bx}\ \Rightarrow\ 30v'_{Ax}+50v'_{Bx}=-160.7$$

Restitution:

$$v'_{Bx}-v'_{Ax}=0.9\,(30+21.21)=46.09$$

Substituting $v'_{Bx}=v'_{Ax}+46.09$:

$$80v'_{Ax}+2304.6=-160.7\ \Rightarrow\ v'_{Ax}=-30.82\ \text{m/s},\qquad v'_{Bx}=15.28\ \text{m/s}$$

Final velocities

Ball A: $v'_A=30.82$ m/s, directed along the negative x-axis (it rebounds straight back, $180^\circ$).

Ball B:

$$v'_B=\sqrt{15.28^{2}+21.21^{2}}=26.14\ \text{m/s},\qquad \theta=\tan^{-1}\frac{21.21}{15.28}=54.2^\circ\ \text{above the positive x-axis}$$

Answer: $v'_A=30.8$ m/s at $180^\circ$; $v'_B=26.1$ m/s at $54.2^\circ$ above the +x direction. (If B instead moves down-left, the sign of B's y-component reverses and its direction becomes $54.2^\circ$ below +x.)

Principle of conservation of energy

The total mechanical energy (kinetic + potential) of a system remains constant when only conservative forces (gravity, spring) do work:

$$T_1+V_1=T_2+V_2$$

Example: a body of mass $m$ dropped from height $h$. At the top $T_1=0$, $V_1=mgh$; just before hitting the ground $V_2=0$, $T_2=\tfrac12mv^{2}$. So $mgh=\tfrac12mv^{2}$ and $v=\sqrt{2gh}$. Potential energy changes completely into kinetic energy. Where friction acts, the energy is not conserved: the work of friction is subtracted.

Block on the incline

Data: incline 3 m high, 4 m horizontal, so length $AB=5$ m, $\cos\theta=4/5=0.8$. $\mu_k=0.3$ (the block slides since $\tan\theta=0.75>\mu_s=0.35$).

 A
 |\
 | \ 5 m
3m|  \
 |   \ theta
 +----B==========C
        friction

A to B (work-energy):

$$mg(3)-\mu_k(mg\cos\theta)(5)=\tfrac12mv_B^{2}$$ $$v_B^{2}=2g\left[3-0.3(0.8)(5)\right]=2(9.81)(1.8)=35.32\ \text{m}^2/\text{s}^2$$ $$v_B=5.943\ \text{m/s}$$

B to C (comes to rest):

$$\tfrac12mv_B^{2}=\mu_k mg\,d\ \Rightarrow\ d=\frac{v_B^{2}}{2\mu_kg}=\frac{35.32}{2(0.3)(9.81)}=6.0\ \text{m}$$

Answer: the block travels $6.0$ m along the horizontal plane before stopping.

Geometry (read from the figure): radius $r=0.127$ m, centre O. Collar released at A (right end of the horizontal diameter), B is the lowest point. The spring is fixed at C, $0.178$ m to the left of the circle, on the horizontal line through O. Take O as origin: A $=(0.127,0)$, B $=(0,-0.127)$, C $=(-0.305,0)$. Undeformed length $0.127$ m, $k=146$ N/m, $m=0.45$ kg.

Spring lengths

$$\begin{aligned} L_A&=0.127+0.305=0.432\ \text{m}\ \Rightarrow\ x_A=0.432-0.127=0.305\ \text{m}\\ L_B&=\sqrt{0.305^{2}+0.127^{2}}=0.3304\ \text{m}\ \Rightarrow\ x_B=0.2034\ \text{m} \end{aligned}$$

Speed at B (conservation of energy, datum at B)

$$mg\,r+\tfrac12kx_A^{2}=\tfrac12mv_B^{2}+\tfrac12kx_B^{2}$$ $$0.45(9.81)(0.127)+\tfrac12(146)(0.305)^{2}=\tfrac12(0.45)v_B^{2}+\tfrac12(146)(0.2034)^{2}$$ $$0.5607+6.7908=0.225v_B^{2}+3.0200\ \Rightarrow\ v_B^{2}=19.25,\ \ v_B=4.388\ \text{m/s}$$

Normal force at B

Spring force $F_s=kx_B=146(0.2034)=29.69$ N, directed from B towards C. Its vertical (upward) component is

$$F_{s,y}=29.69\cdot\frac{0.127}{0.3304}=11.41\ \text{N}$$

The horizontal component is tangential, so it does not enter the normal equation. Newton's second law toward the centre O (upward) at B:

$$N+F_{s,y}-mg=\frac{mv_B^{2}}{r}$$ $$N=\frac{0.45(19.25)}{0.127}+0.45(9.81)-11.41=68.2+4.41-11.41=61.2\ \text{N}$$

Answer: $v_B\approx4.39$ m/s; normal force $N\approx61.2$ N acting upward, towards O.

Assumptions: the arc is smooth; the origin of the x-axis is the point B (the lowest point of the arc, where the 0.4 kg particle rests); $\mu_k=0.3$ acts on the horizontal surface; $g=9.81\ \text{m/s}^2$.

Speed of the 0.5 kg particle at B

Drop in height $h=0.25\sin30^\circ=0.125$ m.

$$\tfrac12m_1u^{2}=m_1gh\ \Rightarrow\ u=\sqrt{2(9.81)(0.125)}=1.566\ \text{m/s}$$

Elastic direct impact ($e=1$)

$m_1=0.5$ kg, $m_2=0.4$ kg (initially at rest). Momentum and restitution:

$$0.5(1.566)=0.5v_1+0.4v_2,\qquad v_2-v_1=1.566$$ $$v_1=\frac{m_1-m_2}{m_1+m_2}u=\frac{0.1}{0.9}(1.566)=0.1740\ \text{m/s}$$ $$v_2=\frac{2m_1}{m_1+m_2}u=\frac{1.0}{0.9}(1.566)=1.740\ \text{m/s}$$

Both move in the +x direction, and $v_2>v_1$, so they do not collide again.

Sliding distances (work-energy with friction)

$$\tfrac12mv^{2}=\mu_kmg\,d\ \Rightarrow\ d=\frac{v^{2}}{2\mu_kg}$$ $$d_1=\frac{0.1740^{2}}{2(0.3)(9.81)}=0.00514\ \text{m},\qquad d_2=\frac{1.740^{2}}{2(0.3)(9.81)}=0.514\ \text{m}$$

Answer: the 0.5 kg particle stops $0.0051$ m (5.1 mm) from the origin and the 0.4 kg particle stops $0.514$ m from the origin, both on the positive x-side.

Coefficient of restitution (direct central impact)

Two particles A and B (masses $m_A,m_B$) move along the same line with $v_A>v_B$ and collide.

Deformation period: they are compressed until they have a common velocity $u$. Let $P$ be the impulse of the compression force.

$$m_Av_A-\int P\,dt=m_Au,\qquad m_Bv_B+\int P\,dt=m_Bu$$

Restitution period: they separate under the restoring impulse $\int R\,dt$:

$$m_Au-\int R\,dt=m_Av'_A,\qquad m_Bu+\int R\,dt=m_Bv'_B$$

Definition: $e=\dfrac{\int R\,dt}{\int P\,dt}$. From the A equations: $\int P\,dt=m_A(v_A-u)$ and $\int R\,dt=m_A(u-v'_A)$, so $e=\dfrac{u-v'_A}{v_A-u}$. From the B equations: $e=\dfrac{v'_B-u}{u-v_B}$. Adding numerators and denominators of these equal ratios:

$$e=\frac{v'_B-v'_A}{v_A-v_B}$$

i.e. velocity of separation divided by velocity of approach; $e=1$ elastic, $e=0$ plastic.

Car on the track

$W=30$ kN, starts from rest at 1, which is 40 m above point 2. Point 3 is 15 m above point 2, so 25 m below point 1. The track is frictionless, so energy is conserved.

 1 o
    \                    3 (crest)
     \                  _o_
      \               /     \
       \__________ 2 /
       rho2 = 20 m

Force at point 2 (lowest point):

$$v_2^{2}=2g(40)=80g$$ $$N-W=\frac{W}{g}\frac{v_2^{2}}{\rho_2}\ \Rightarrow\ N=W\left(1+\frac{80}{20}\right)=5W=5(30)=150\ \text{kN}$$

Minimum radius at crest 3: speed there from 1 to 3 (drop 25 m): $v_3^{2}=2g(25)=50g$. For the car to stay on the track the normal force $N_3\ge0$:

$$W-N_3=\frac{W}{g}\frac{v_3^{2}}{\rho_3}\ \Rightarrow\ \rho_3\ge\frac{v_3^{2}}{g}=\frac{50g}{g}=50\ \text{m}$$

Answer: force of the track on the car at 2 is $150$ kN (upward); minimum safe radius of curvature at 3 is $\rho_3=50$ m.

Assumed geometry (figure not shown): the rod is a quarter circle of radius $R=0.6$ m, so the slider starts at A at the level of the circle centre and drops $h=0.6$ m to the lowest point B, where it meets the horizontal spring ($K=20$ kN/m). $m=4$ kg, rod smooth.

a) Velocity at the spring

Only gravity does work from A to B:

$$mgh=\tfrac12mv^{2}\ \Rightarrow\ v=\sqrt{2gh}=\sqrt{2(9.81)(0.6)}=3.431\ \text{m/s}$$

b) Maximum deflection of the spring

All kinetic energy at B is stored in the spring when the slider stops (horizontal compression, no further drop):

$$\tfrac12mv^{2}=\tfrac12Kx^{2}$$ $$x=v\sqrt{\frac{m}{K}}=3.431\sqrt{\frac{4}{20000}}=0.0485\ \text{m}$$

Answer: $v\approx3.43$ m/s; maximum spring deflection $x\approx48.5$ mm. (If the slider moves down an additional distance $x$ while compressing the spring, use $mg(h+x)=\tfrac12Kx^{2}$.)