Resolving Forces and Inclined Planes: past paper questions & worked solutions

Take the initial direction of P as positive. Then before impact, $v_P=3u$ and $v_Q=-4u$. The impulse on P acts opposite to P\'s initial motion, so its signed impulse is $-\dfrac{33}{2}mu$.

For P, impulse equals change in momentum: $-\dfrac{33}{2}mu=3m(v_P-3u)$. Dividing by $3m$ gives $v_P-3u=-\dfrac{11u}{2}$, so $v_P=-\dfrac{5u}{2}$.

The negative sign means P has reversed direction. Its speed is therefore $\left|-\dfrac{5u}{2}\right|=\dfrac{5u}{2}$.

Q receives impulse $+\dfrac{33}{2}mu$ in the positive direction. Hence $\dfrac{33}{2}mu=5m(v_Q-(-4u))=5m(v_Q+4u)$.

Dividing by $5m$ gives $v_Q+4u=\dfrac{33u}{10}$, so $v_Q=-\dfrac{7u}{10}$. Q is still moving in its original direction, and its speed is $\dfrac{7u}{10}$.

The bucket and cement have total mass $15+25=40$ kg and acceleration $0.2\ \mathrm{m\,s^{-2}}$ upward.

Taking upward as positive: $T-40g=40(0.2)$.

$T-392=8$, so $T=400\ \mathrm{N}$.

The bucket and tools are moving downward but decelerating, so their acceleration is upward with magnitude $0.1\ \mathrm{m\,s^{-2}}$.

For the $12$ kg box, upward force is the normal reaction $R$, and downward force is weight $12g$.

$R-12g=12(0.1)$. Thus $R=117.6+1.2=118.8\ \mathrm{N}$, about $119\ \mathrm{N}$.

For the trailer, take up the road as positive. The forces along the road are tension $2060$ N up the road, resistance $300$ N down the road, and weight component $400g\sin\alpha$ down the road.

$2060-300-400g\sin\alpha=400a$. Since $\sin\alpha=\dfrac17$, this becomes $2060-300-400g\cdot\dfrac17=400a$.

Using $g=9.8$, $400g\cdot\dfrac17=560$. Hence $1200=400a$ and $a=3\ \mathrm{m\,s^{-2}}$.

For the car: $D-420-800g\sin\alpha-2060=800a$.

$D-420-800g\cdot\dfrac17-2060=800(3)$. Since $800g\cdot\dfrac17=1120$, $D=420+1120+2060+2400=6000$ N.

$\text{Impulse} = \text{Final momentum} - \text{Initial momentum}$ For particle Q (mass = 3m): ; Initial velocity: $-u$ (moving left, negative) ; Final velocity: $+2w$ (moving right, positive) $\text{Impulse} = 3m(2w) - 3m(-u)$ $\frac{7mu}{2} = 6mw + 3mu$ $\frac{7mu}{2} = 3m(2w + u)$ $\frac{7u}{2} = 6w + 3u$ $6w = \frac{7u}{2} - 3u$ $6w = \frac{7u - 6u}{2}$ $6w = \frac{u}{2}$ $w = \frac{u}{12}$

$\text{Total momentum before} = \text{Total momentum after}$ Before collision: $\text{P momentum} = 7m \times ku = 7mku \text{ (right)}$ $\text{Q momentum} = 3m \times (-u) = -3mu \text{ (left)}$ $\text{Total} = 7mku - 3mu$ After collision: $\text{P momentum} = 7m \times w = 7mw \text{ (right)}$ $\text{Q momentum} = 3m \times 2w = 6mw \text{ (right)}$ $\text{Total} = 7mw + 6mw = 13mw$

$7mku - 3mu = 13mw$ $7ku - 3u = 13w$

$7ku - 3u = 13 \times \frac{u}{12}$ $7ku - 3u = \frac{13u}{12}$ $7ku = \frac{13u}{12} + 3u$ $7ku = \frac{13u + 36u}{12}$ $7ku = \frac{49u}{12}$ $7k = \frac{49}{12}$ $k = \frac{49}{84} = \frac{7}{12}$

P could move either right (velocity = $+w$) or left (velocity = $-w$) Case 1: P moves right with velocity $+w$ Already solved: $k = \frac{7}{12}$ Case 2: P moves left with velocity $-w$ Total momentum after = $-7mw + 6mw = -mw$ $7mku - 3mu = -mw$ $7ku - 3u = -w$ $7ku - 3u = -\frac{u}{12}$ $7ku = 3u - \frac{u}{12}$ $7ku = \frac{36u - u}{12}$ $7ku = \frac{35u}{12}$ $k = \frac{35}{84} = \frac{5}{12}$

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

The force $X$ acts along the plane, so the normal reaction is $R=mg\cos\alpha=\dfrac45mg$.

The limiting friction is $F=\mu R=\dfrac45\mu mg$.

When $X=2P$, the particle is about to slide up, so friction acts down the plane. Hence $2P=mg\sin\alpha+F=\dfrac35mg+\dfrac45\mu mg$.

When $X=P$, the particle is about to slide down, so friction acts up the plane. Hence $P+F=mg\sin\alpha$, so $P=\dfrac35mg-\dfrac45\mu mg$.

Double the second equation: $2P=\dfrac65mg-\dfrac85\mu mg$. This equals the first expression for $2P$.

$\dfrac65mg-\dfrac85\mu mg=\dfrac35mg+\dfrac45\mu mg$. Cancel $mg$ and multiply by $5$: $6-8\mu=3+4\mu$. Hence $3=12\mu$ and $\mu=\dfrac14$.

Vertically, $R_C+R_D=50g+25g=75g$. Given $2R_D=3R_C$, we have $R_D=1.5R_C$. Therefore $2.5R_C=75g$, so $R_C=30g$ and $R_D=45g$.

$D$ is $6-1.8=4.2$ m from A. The support C is $4.2-0.9=3.3$ m to the left of D, the child at E is $4.2-3=1.2$ m to the left of D, and the centre of mass is $x$ m to the left of D.

Balancing moments about D gives $50gx+25g(1.2)=R_C(3.3)$.

Substitute $R_C=30g$: $50x+30=99$, so $50x=69$ and $x=\dfrac{69}{50}=1.38$.

When the block at B is on the point of making the beam tilt, the beam is about to pivot about D and the reaction at C is zero.

The child and the beam act to the left of D, and the block at B acts $1.8$ m to the right of D. Hence $25g(1.2)+50g(1.38)=Mg(1.8)$.

$30+69=1.8M$, so $M=\dfrac{99}{1.8}=55$.

The beam length is $8a$. C is at distance $a$ from A, and D is at distance $6a$ from A because $DB=2a$.

Let the reaction at C be $R$. The reaction at D is $3R$.

$R+3R=60g$, so $4R=60g$ and $R=15g$.

The beam's weight acts at distance $x$ from A. Hence $60g x=R(a)+3R(6a)$.

Substitute $R=15g$: $60g x=15ga+45g(6a)=285ga$. Thus $x=\dfrac{285a}{60}=\dfrac{19a}{4}$.

When the box is added, let each support reaction be $S$. Vertical equilibrium gives $2S=(60+M)g$.

The box is at $AE=2a$. Moments about A give $60g\left(\dfrac{19a}{4}\right)+Mg(2a)=S(a)+S(6a)=7aS$.

Substitute $S=\dfrac{(60+M)g}{2}$ and cancel $ag$: $285+2M=\dfrac{7}{2}(60+M)=210+\dfrac72M$. Hence $75=\dfrac32M$ and $M=50$.

Given: $\tan\theta = \frac{3}{4}$ Using 3-4-5 right triangle: $\sin\theta = \frac{3}{5}, \quad \cos\theta = \frac{4}{5}$

$\perp: \sum F = 0$ $R = mg\cos\theta$ $R = \frac{4mg}{5}$

Package moving UP, so friction acts DOWN the plane: $F = \mu R$ $= \frac{1}{4} \times \frac{4mg}{5}$ $= \frac{mg}{5}$

Taking up the plane as positive: $\parallel: \sum F = ma$ $-mg\sin\theta - F = ma$ $-mg \times \frac{3}{5} - \frac{mg}{5} = ma$ $-\frac{3mg}{5} - \frac{mg}{5} = ma$ $-\frac{4mg}{5} = ma$ $a = -\frac{4g}{5}$ Deceleration = $\left|-\frac{4g}{5}\right| = \frac{4g}{5}$

Package projected up plane, comes to rest at B: ; $u = U$ m s$^{-1}$ (unknown) ; $v = 0$ m s$^{-1}$ (comes to rest) ; $a = -\frac{4g}{5}$ m s$^{-2}$ (from part a) ; $s = 1.5$ m

$v^2 = u^2 + 2as$ $0^2 = U^2 + 2\left(-\frac{4g}{5}\right)(1.5)$ $0 = U^2 - 2 \times \frac{4 \times 9.8}{5} \times 1.5$ $0 = U^2 - \frac{8 \times 9.8 \times 1.5}{5}$ $0 = U^2 - 23.52$ $U^2 = 23.52$ $U = 4.85 \text{ m s}^{-1}$ Or using exact form: $U = \sqrt{2 \times \frac{4g}{5} \times 1.5} = \sqrt{\frac{12g}{5}}$

Package held at rest on slope by force X up the plane. Without X, package would slide down (component down > friction up). For MINIMUM X, friction acts UP the plane (at limiting friction).

Mass = 2 kg, so weight = $2g$ N $R = 2g\cos\theta$ $= 2g \times \frac{4}{5}$ $= \frac{8g}{5} \text{ N}$

Friction acts UP the plane (limiting): $F = \mu R$ $= \frac{1}{4} \times \frac{8g}{5}$ $= \frac{2g}{5} \text{ N}$

For equilibrium (package at rest): $\parallel: \sum F = 0$ Taking up plane as positive: $X + F - 2g\sin\theta = 0$ $X + \frac{2g}{5} - 2g \times \frac{3}{5} = 0$ $X + \frac{2g}{5} - \frac{6g}{5} = 0$ $X = \frac{6g}{5} - \frac{2g}{5}$ $X = \frac{4g}{5}$ $X = \frac{4 \times 9.8}{5}$ $X = 7.84 \text{ N}$

For car plus trailer, the total mass is $1800$ kg and the total resistance is $2R+R=3R$. Taking up the slope as positive: $3000-1800g\sin\alpha-3R=1800(0.75)$.

Since $\sin\alpha=\dfrac1{12}$, $1800g\sin\alpha=1800\cdot9.8\cdot\dfrac1{12}=1470$. So $3000-1470-3R=1350$, giving $180=3R$ and $R=60\ \mathrm{N}$.

For the trailer while it is being towed, $T-600g\sin\alpha-R=600(0.75)$.

$T-600\cdot9.8\cdot\dfrac1{12}-60=450$, so $T-490-60=450$ and $T=1000\ \mathrm{N}$.

After the tow bar breaks, the trailer is still moving up the slope, but the component of weight and resistance act down the slope. Taking up the slope as positive: $-600g\sin\alpha-60=600a$.

$-490-60=600a$, so $a=-\dfrac{550}{600}=-\dfrac{11}{12}\ \mathrm{m\,s^{-2}}$.

With initial speed $12\ \mathrm{m\,s^{-1}}$ and final speed $0$, use $v^2=u^2+2as$: $0=12^2+2\left(-\dfrac{11}{12}\right)d$. Hence $d=\dfrac{864}{11}=78.5\ldots\ \mathrm{m}$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

If there is no friction, the component of the horizontal force up the plane balances the component of weight down the plane: $X\cos\alpha=2g\sin\alpha$.

$14.7\cdot\dfrac45=11.76$, and $2(9.8)\cdot\dfrac35=11.76$. The components balance exactly, so the frictional force is zero.

For the smallest possible $X$, the particle is about to slide down the plane, so friction acts up the plane.

The horizontal force increases the normal reaction, so $R=2g\cos\alpha+X\sin\alpha$.

With $\mu=0.5$, the limiting friction is $F=0.5R=0.5(2g\cos\alpha+X\sin\alpha)$.

At the limiting case, $X\cos\alpha+F=2g\sin\alpha$.

Substitute the exact trig values: $\dfrac45X+0.5\left(2g\cdot\dfrac45+X\cdot\dfrac35\right)=2g\cdot\dfrac35$. This gives $\dfrac{11}{10}X+\dfrac{4g}{5}=\dfrac{6g}{5}$, so $X=\dfrac{4g}{11}=3.56\ldots\ \mathrm{N}$.

On a horizontal surface, $R=mg$ and the friction is $F=\mu R=\dfrac17mg$. The friction is the only horizontal force, so the acceleration is opposite the motion with magnitude $\dfrac{F}{m}=\dfrac{g}{7}=1.4\ \mathrm{m\,s^{-2}}$.

From A to B, $u=U$, $v=\dfrac{U}{2}$, $a=-\dfrac{g}{7}$ and $t=0.75$. Using $v=u+at$: $\dfrac{U}{2}=U-\dfrac{g}{7}(0.75)$.

With $g=9.8$, $\dfrac{g}{7}(0.75)=1.05$, so $\dfrac{U}{2}=U-1.05$. Hence $U=2.10\ \mathrm{m\,s^{-1}}$, which is $2.1\ \mathrm{m\,s^{-1}}$ to 2 significant figures.

From B to C, the speed falls from $\dfrac{U}{2}$ to $\dfrac{U}{3}$ with the same acceleration $-1.4$. Thus $\dfrac{U}{3}=\dfrac{U}{2}-1.4t$. With $U=2.1$, this gives $0.7=1.05-1.4t$, so $t=0.25\ \mathrm{s}$.

Use average speed on each section. $AB=\dfrac{U+U/2}{2}(0.75)=1.18125\ \mathrm{m}$ and $BC=\dfrac{U/2+U/3}{2}(0.25)=0.21875\ \mathrm{m}$. Therefore $AC=1.18125+0.21875=1.4\ \mathrm{m}$.

P starts from rest and immediately after the impulse has speed $5\ \mathrm{m\,s^{-1}}$. Since the mass of P is $3m$, $\lambda=3m(5-0)=15m$.

Between the impulse and the collision, P slows from $5$ to $2.5\ \mathrm{m\,s^{-1}}$ over $1.4$ m. Using $v^2=u^2+2as$: $(2.5)^2=5^2+2a(1.4)$.

$6.25=25+2.8a$, so $a=\dfrac{-18.75}{2.8}=-6.696\ldots\ \mathrm{m\,s^{-2}}$.

On the horizontal surface, $R=3mg$ and friction is $F=\mu R=3\mu mg$. This friction is opposite the motion, so $-3\mu mg=3ma$ and therefore $a=-\mu g$.

$\mu=\dfrac{-a}{g}=\dfrac{6.696\ldots}{9.8}=0.683\ldots$, so $\mu=0.68$ to 2 significant figures.

Immediately before collision, P has speed $2.5$ and Q is at rest. Immediately after, Q has speed $2.1$ and P has speed $v$. Conservation of momentum gives $3m(2.5)+m(0)=3mv+m(2.1)$.

$7.5=3v+2.1$, so $3v=5.4$ and $v=1.8\ \mathrm{m\,s^{-1}}$.

P is on the point of moving up the plane, so friction acts down the plane. The tension has component $0.75mg\cos\alpha$ up the plane, and the weight component is $mg\sin\alpha$ down the plane.

For equilibrium parallel to the plane, $0.75mg\cos\alpha=mg\sin\alpha+F$. Hence $F=0.75mg\cos\alpha-mg\sin\alpha$.

The string has a component away from the plane, so $R=mg\cos\alpha-0.75mg\sin\alpha$.

Given $\mu=\dfrac12$, limiting friction gives $F=\dfrac12R$. Therefore $0.75mg\cos\alpha-mg\sin\alpha=\dfrac12(mg\cos\alpha-0.75mg\sin\alpha)$.

Divide by $mg$ and collect terms: $0.75\cos\alpha-\sin\alpha=0.5\cos\alpha-0.375\sin\alpha$. Thus $0.25\cos\alpha=0.625\sin\alpha$, so $\tan\alpha=\dfrac{0.25}{0.625}=\dfrac25$.

Without the string, $R=mg\cos\alpha$ and the maximum friction is $\dfrac12mg\cos\alpha$. The component of weight down the plane is $mg\sin\alpha$.

Since $\tan\alpha=\dfrac25<\dfrac12$, we have $mg\sin\alpha<\dfrac12mg\cos\alpha$. The available friction is large enough to hold P, so P remains at rest.

From $\tan\alpha=\dfrac34$, use $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

The horizontal force has a component into the plane, so $R=5g\cos\alpha+H\sin\alpha=4g+\dfrac35H$.

For the smallest $H$, the particle is on the point of sliding down, so friction acts up the plane. Parallel to the plane, $H\cos\alpha+F=5g\sin\alpha$.

With $F=\dfrac14R$, the parallel equation becomes $\dfrac45H+\dfrac14\left(4g+\dfrac35H\right)=3g$. Hence $0.95H+g=3g$, so $H=\dfrac{40g}{19}=20.6\ \mathrm{N}$.

When $H$ is removed, $R=5g\cos\alpha=4g$ and friction is $\dfrac14R=g$. The component of weight down the plane is $5g\sin\alpha=3g$, so the resultant down the plane is $2g$.

$2g=5a$, so $a=\dfrac{2g}{5}=3.92\ \mathrm{m\,s^{-2}}$ down the plane.

From rest, the particle moves $1.5$ m with acceleration $2g/5$. Using $s=\dfrac12at^2$: $1.5=\dfrac12\left(\dfrac{2g}{5}\right)T^2=\dfrac{g}{5}T^2$. Thus $T=\sqrt{\dfrac{7.5}{9.8}}=0.875\ \mathrm{s}$.

Given $\tan\theta=\dfrac34$, use $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$.

The box moves at constant speed, so it is in equilibrium. Vertically, $R+T\sin\theta=mg$, hence $R=mg-T\sin\theta$.

Horizontally, the tension component balances friction: $T\cos\theta=F$.

With $\mu=\dfrac13$, $F=\dfrac13R$. Therefore $T\cos\theta=\dfrac13(mg-T\sin\theta)$.

Substitute $\sin\theta=\dfrac35$ and $\cos\theta=\dfrac45$: $\dfrac45T=\dfrac13\left(mg-\dfrac35T\right)$. This gives $\dfrac45T=\dfrac{mg}{3}-\dfrac15T$, so $T=\dfrac{mg}{3}$.

On the horizontal plane without the rope, $R=mg$ and friction is $\dfrac13mg$. Hence the acceleration is $-\dfrac{g}{3}$.

When the speed has fallen from $u$ to $\dfrac12u$, use $v^2=u^2+2as$: $\left(\dfrac{u}{2}\right)^2=u^2+2\left(-\dfrac{g}{3}\right)d$.

$\dfrac{u^2}{4}=u^2-\dfrac{2gd}{3}$, so $\dfrac{2gd}{3}=\dfrac{3u^2}{4}$ and $d=\dfrac{9u^2}{8g}$.

Given $\tan\alpha=\dfrac{5}{12}$, use $\sin\alpha=\dfrac{5}{13}$ and $\cos\alpha=\dfrac{12}{13}$.

Q has mass $0.5m$ and the system is in equilibrium, so the tension is $T=0.5mg$.

P is on the point of slipping up the plane, so friction acts down the plane. Along the plane, $T=mg\sin\alpha+F$.

$F=T-mg\sin\alpha=\dfrac12mg-\dfrac{5}{13}mg=\dfrac{3}{26}mg$.

$R=mg\cos\alpha=\dfrac{12}{13}mg$. Since $F=\mu R$, $\mu=\dfrac{3mg/26}{12mg/13}=\dfrac18$.

P moves down the plane, so friction acts up the plane. The resultant down the plane is $mg\sin\alpha-F=\dfrac{5}{13}mg-\dfrac18\cdot\dfrac{12}{13}mg=\dfrac{7}{26}mg$.

As P has mass $m$, $a=\dfrac{7g}{26}$ down the plane.

From rest over $0.8$ m, $V^2=2\left(\dfrac{7g}{26}\right)(0.8)$. With $g=9.8$, $V=2.05\ \mathrm{m\,s^{-1}}$ to 3 significant figures.

The string makes angle $\theta$ with the vertical rod and $\tan\theta=\dfrac{12}{5}$. Hence $\sin\theta=\dfrac{12}{13}$ and $\cos\theta=\dfrac{5}{13}$.

The upward component of the tension is $10\cos\theta=10\cdot\dfrac{5}{13}=\dfrac{50}{13}$ N. The weight is $0.2g=1.96$ N.

Since the upward component of tension is larger than the weight, friction must act downwards. Its magnitude is $\dfrac{50}{13}-1.96=1.89\ldots\ \mathrm{N}$, so $F=1.9\ \mathrm{N}$.

For variable tension $T$, the normal reaction from the rod is produced by the horizontal component of tension: $R=T\sin\theta=\dfrac{12T}{13}$.

Since $\mu=\dfrac14$, limiting friction is $F=\mu R=\dfrac14\cdot\dfrac{12T}{13}=\dfrac{3T}{13}$.

For the minimum tension, the ring would tend to slide down, so friction acts upwards. Vertical equilibrium gives $T\cos\theta+F=0.2g$, hence $\dfrac{5T}{13}+\dfrac{3T}{13}=1.96$. Thus $T=3.185\ldots\ \mathrm{N}$, so $T_{\min}=3.2\ \mathrm{N}$.

For the maximum tension, the ring would tend to move up, so friction acts downwards. Vertical equilibrium gives $T\cos\theta=0.2g+F$, hence $\dfrac{5T}{13}=1.96+\dfrac{3T}{13}$. Thus $T=12.74\ldots\ \mathrm{N}$, so $T_{\max}=13\ \mathrm{N}$.

The particle is on the point of sliding down the plane, so friction acts up the plane. Also $X=mg$.

Both the weight $kmg$ and the horizontal force $mg$ press the particle into the plane. Therefore $R=kmg\cos\theta+mg\sin\theta$.

Down the plane is the component $kmg\sin\theta$. Up the plane are friction $F$ and the component $mg\cos\theta$. Thus $F+mg\cos\theta=kmg\sin\theta$.

So $F=kmg\sin\theta-mg\cos\theta$.

$\mu=\dfrac{F}{R}=\dfrac{kmg\sin\theta-mg\cos\theta}{kmg\cos\theta+mg\sin\theta}$.

Cancel $mg$ and divide numerator and denominator by $\cos\theta$: $\mu=\dfrac{k\tan\theta-1}{k+\tan\theta}$, as required.

When $k=1$, $\mu=\dfrac{\tan\theta-1}{1+\tan\theta}$. Since $\mu>0$ and the denominator is positive, $\tan\theta-1>0$.

Therefore $\tan\theta>1$, so because $\theta$ is acute, $\theta>45^\circ$.

$\uparrow: \sum F_{\text{vertical}} = 0$ $R + P\sin 30° = mg$ $R + 0.5P = 4 \times 9.8$ $R = 39.2 - 0.5P \quad \text{...(1)}$

At limiting equilibrium (on point of sliding): $F = \mu R$ $F = \frac{2}{3} \times R$ $F = \frac{2}{3}(39.2 - 0.5P)$ $F = 26.133... - \frac{P}{3} \quad \text{...(2)}$

Box on point of sliding means $a = 0$: $\rightarrow: \sum F_{\text{horizontal}} = 0$ $P\cos 30° = F$ $P \times \frac{\sqrt{3}}{2} = 26.133... - \frac{P}{3}$ $0.866P = 26.133 - 0.333P$ $0.866P + 0.333P = 26.133$ $1.199P = 26.133$ $P = 21.8 \text{ N (3 s.f.)}$

From equation (1) in part (a): $R = 39.2 - 0.5P$ $R = 39.2 - 0.5(25)$ $R = 39.2 - 12.5$ $R = 26.7 \text{ N}$

Box is now moving, so: $F = \mu R$ $F = \frac{2}{3} \times 26.7$ $F = 17.8 \text{ N}$

$\rightarrow: \sum F = ma$ $P\cos 30° - F = ma$ $25 \times 0.866 - 17.8 = 4a$ $21.65 - 17.8 = 4a$ $3.85 = 4a$ $a = 0.9625$ $a = 0.96 \text{ m s}^{-2} \text{ (2 s.f.)}$

$u = 0$ m s$^{-1}$ (starts from rest) ; $a = 0.9625$ m s$^{-2}$ (from part b) ; $s = 1.5$ m ; $v = ?$

$v^2 = u^2 + 2as$ $v^2 = 0^2 + 2(0.9625)(1.5)$ $v^2 = 2.8875$ $v = \sqrt{2.8875}$ $v = 1.699...$ $v = 1.7 \text{ m s}^{-1} \text{ (2 s.f.)}$

Given $\tan\alpha=\dfrac{7}{24}$, use $\sin\alpha=\dfrac{7}{25}$ and $\cos\alpha=\dfrac{24}{25}$.

For block A, $R=mg\cos\alpha=\dfrac{24mg}{25}$. Since $\mu=\dfrac13$, friction is $F=\dfrac13R=\dfrac{8mg}{25}$.

A moves up the plane, so friction and the component of weight act down the plane. Taking up the plane as positive: $T-mg\sin\alpha-F=ma$.

Substitute $T=\dfrac43mg$, $mg\sin\alpha=\dfrac{7mg}{25}$ and $F=\dfrac{8mg}{25}$: $ma=\dfrac43mg-\dfrac{15mg}{25}=\left(\dfrac43-\dfrac35\right)mg=\dfrac{11mg}{15}$. Hence $a=\dfrac{11g}{15}$.

Block B moves downwards with the same acceleration. Taking downward as positive: $kmg-T=kma$.

Substitute $T=\dfrac43mg$ and $a=\dfrac{11g}{15}$: $kmg-\dfrac43mg=k m\dfrac{11g}{15}$. Divide by $mg$ to get $k-\dfrac43=\dfrac{11k}{15}$, so $15k-20=11k$ and $k=5$.

The pulley is pulled by two tensions of magnitude $T$. One is vertical and the other is along the plane. The horizontal component of the resultant is $T\cos\alpha$ and the vertical component is $T+T\sin\alpha$.

The magnitude is $T\sqrt{\cos^2\alpha+(1+\sin\alpha)^2}$. With $\sin\alpha=\dfrac{7}{25}$ and $\cos\alpha=\dfrac{24}{25}$, this is $T\sqrt{\dfrac{576}{625}+\dfrac{1024}{625}}=\dfrac{8T}{5}$. Since $T=\dfrac43mg$, the resultant is $\dfrac{32}{15}mg$.

For Q on the horizontal table, $R=mg$, so the friction is $F=\mu mg$.

P moves downward and Q moves towards the pulley. For P, taking downward as positive: $mg-T=ma$.

For Q, taking motion towards the pulley as positive: $T-\mu mg=ma$.

From P, $a=g(1-k)$. From Q, $a=g(k-\mu)$.

Equate the two accelerations: $1-k=k-\mu$. Hence $2k=1+\mu$ and $k=\dfrac{1+\mu}{2}$.

$a=g(1-k)=g\left(1-\dfrac{1+\mu}{2}\right)=\dfrac{g(1-\mu)}{2}$.

Q starts from rest and travels distance $d$, so $d=\dfrac12at^2=\dfrac12\cdot\dfrac{g(1-\mu)}{2}t^2$.

$d=\dfrac{g(1-\mu)t^2}{4}$, so $t=\sqrt{\dfrac{4d}{g(1-\mu)}}$.

If $\mu\ge1$, the maximum available friction on Q is at least $mg$, so it can balance the pull caused by the hanging particle. The system would not start moving.

Q lies on a smooth horizontal plane. The only horizontal force on Q is the tension, so $T=ma$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

For P, $R=4mg\cos\alpha=4mg\cdot\dfrac45=\dfrac{16mg}{5}$. With coefficient $\dfrac14$, friction is $F=\dfrac14R=\dfrac{4mg}{5}$.

P moves down the plane. Taking down the plane as positive: $4mg\sin\alpha-T-F=4ma$.

$4mg\cdot\dfrac35-T-\dfrac{4mg}{5}=4ma$, so $\dfrac{8mg}{5}-T=4ma$.

From Q, $a=\dfrac{T}{m}$. Substitute into the equation for P: $\dfrac{8mg}{5}-T=4T$.

$\dfrac{8mg}{5}=5T$, so $T=\dfrac{8}{25}mg$.

The pulley is pulled by two tensions of magnitude $T$: one horizontal and one along the inclined string. Their resultant has components $T-T\cos\alpha$ horizontally and $T\sin\alpha$ vertically.

Magnitude $=\sqrt{(T-T\cos\alpha)^2+(T\sin\alpha)^2}=T\sqrt{(1-\tfrac45)^2+(\tfrac35)^2}=\dfrac{T}{5}\sqrt{10}$. With $T=\dfrac{8}{25}mg$, the resultant is $\dfrac{8\sqrt{10}}{125}mg$.

The string is light, so the tension can be treated as the same on the P side and the Q side of the pulley.

Since $\tan\theta=\dfrac5{12}$, $\sin\theta=\dfrac5{13}$ and $\cos\theta=\dfrac{12}{13}$.

$R=6g\cos\theta=6g\cdot\dfrac{12}{13}=\dfrac{72g}{13}$.

With $\mu=\dfrac14$, $F=\dfrac14R=\dfrac{18g}{13}=13.6\ \mathrm{N}$ approximately.

While moving up, both friction and the component of weight act down the plane. Taking up the plane as positive: $6a=-6g\sin\theta-F$.

$6a=-6g\cdot\dfrac5{13}-\dfrac{18g}{13}=-\dfrac{48g}{13}$, so $a=-\dfrac{8g}{13}$.

Use $v^2=u^2+2as$ with $u=5$ and $v=0$: $0=25+2\left(-\dfrac{8g}{13}\right)AB$. Hence $AB=2.07\ldots\ \mathrm{m}$.

When sliding down, friction acts up the plane. Taking down the plane as positive: $6a=6g\sin\theta-F=\dfrac{30g}{13}-\dfrac{18g}{13}=\dfrac{12g}{13}$. Thus $a=\dfrac{2g}{13}$.

Starting from rest at B and reaching speed $5$ at C: $25=2\left(\dfrac{2g}{13}\right)BC$. Hence $BC=8.29\ldots\ \mathrm{m}$.

Since $\tan\theta=\dfrac43$, use a $3$-$4$-$5$ triangle: $\sin\theta=\dfrac45$ and $\cos\theta=\dfrac35$.

With one $0.2$ kg block in the bucket, the bucket side has mass $0.45+0.2=0.65$ kg. Since the package is on the point of slipping down the plane, friction on P acts up the plane.

The tension equals the bucket weight, so $T=0.65g$. Along the plane for P in equilibrium, $T+F=g\sin\theta$. Hence $0.65g+F=\dfrac45g$, so $F=\dfrac{3g}{20}$.

Perpendicular to the plane, $R=g\cos\theta=\dfrac35g$. Since $F=\mu R$, $\mu=\dfrac{F}{R}=\dfrac{3g/20}{3g/5}=\dfrac14$.

When P is about to move up the plane, friction acts down the plane. With $n$ blocks in the bucket, the bucket-side mass is $0.45+0.2n$. The bucket weight must exceed the downslope resistance: $(0.45+0.2n)g>g\sin\theta+\mu R$.

Using $g\sin\theta=\dfrac45g$ and $\mu R=\dfrac14\cdot\dfrac35g=\dfrac{3g}{20}$, the condition is $0.45+0.2n>\dfrac45+\dfrac{3}{20}=0.95$. Hence $0.2n>0.50$, so $n>2.5$ and the least integer is $n=3$.

For $n=3$, the bucket-side mass is $0.45+0.6=1.05$ kg and the total moving mass is $1+1.05=2.05$ kg. The driving force is $1.05g-\dfrac45g-\dfrac{3g}{20}=0.10g$.

Using $F=ma$, $0.10g=2.05a$, so $a=\dfrac{0.10g}{2.05}=\dfrac{2g}{41}=0.478\ \mathrm{m\,s^{-2}}$ to 3 significant figures.

Forces on Q (mass = 5m): Taking downward as positive for Q: $\text{For Q: } \sum F = ma$ $5mg - T = 5ma$

$\perp \text{ to plane: } \sum F = 0$ $R = 2mg\cos\alpha$ $R = 2mg \times \frac{12}{13}$ $R = \frac{24mg}{13}$

P is moving UP the plane, so friction acts DOWN the plane: $F = \mu R$ $= \frac{1}{8} \times \frac{24mg}{13}$ $= \frac{3mg}{13}$

Taking up the plane as positive: $\parallel \text{ to plane: } \sum F = ma$ $T - 2mg\sin\alpha - F = 2ma$ $T - 2mg \times \frac{5}{13} - \frac{3mg}{13} = 2ma$ $T - \frac{10mg}{13} - \frac{3mg}{13} = 2ma$ $T - \frac{13mg}{13} = 2ma$ $T - mg = 2ma \quad \text{...(1)}$

From part (a): $5mg - T = 5ma$ ...(2) Add equations (1) and (2): $(T - mg) + (5mg - T) = 2ma + 5ma$ $4mg = 7ma$ $a = \frac{4g}{7}$

From equation (1): $T = mg + 2ma$ $= mg + 2m \times \frac{4g}{7}$ $= mg + \frac{8mg}{7}$ $= \frac{7mg + 8mg}{7}$ $= \frac{15mg}{7}$

An inextensible string means the string cannot stretch. This has the following consequence: For an inextensible string over a pulley: The magnitude of acceleration of both connected particles must be equal. In this problem: $|a_P| = |a_Q| = a$ How this was used in part (b): When we wrote the equations of motion: ; For P: $T - mg = 2ma$ ; For Q: $5mg - T = 5ma$ We used the same acceleration $a$ for both particles because the string is inextensible. If the string could stretch, the two particles could have different accelerations.

The string from Q is vertical (downward). The string from P is along the plane at angle $\alpha$ to horizontal. Angle between the two strings = $90^\circ + \alpha$

Let's take horizontal (right) and vertical (up) as positive directions. Tension from Q: ; Horizontal component: $0$ ; Vertical component: $-T = -\frac{15mg}{7}$ (downward) Tension from P along plane: The plane makes angle $\alpha$ with horizontal, so: ; Horizontal component: $-T\cos\alpha = -\frac{15mg}{7} \times \frac{12}{13} = -\frac{180mg}{91}$ ; Vertical component: $T\sin\alpha = \frac{15mg}{7} \times \frac{5}{13} = \frac{75mg}{91}$

$R_x = 0 - \frac{180mg}{91} = -\frac{180mg}{91}$ $R_y = -\frac{15mg}{7} + \frac{75mg}{91}$ Convert $\frac{15mg}{7}$ to denominator 91: $\frac{15mg}{7} = \frac{195mg}{91}$ $R_y = -\frac{195mg}{91} + \frac{75mg}{91} = -\frac{120mg}{91}$

$|R| = \sqrt{R_x^2 + R_y^2}$ $= \sqrt{\left(\frac{180mg}{91}\right)^2 + \left(\frac{120mg}{91}\right)^2}$ $= \frac{mg}{91}\sqrt{180^2 + 120^2}$ $= \frac{mg}{91}\sqrt{32400 + 14400}$ $= \frac{mg}{91}\sqrt{46800}$ $= \frac{mg}{91} \times 216.333...$ $= 2.377... mg$ $= 2.38mg \text{ (3 s.f.)}$ Therefore: $k = 2.38$

A moves up the plane, so friction and the component of weight act down the plane. Taking up the plane as positive: $T-2mg\sin\alpha-F=2ma$.

B moves downward. Taking downward as positive: $3mg-T=3ma$.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$. The normal reaction on A is $R=2mg\cos\alpha=\dfrac{8mg}{5}$, so $F=\dfrac12R=\dfrac{4mg}{5}$.

$T-2mg\cdot\dfrac35-\dfrac{4mg}{5}=2ma$, so $T-2mg=2ma$.

From B, $T=3mg-3ma$. Substitute into $T-2mg=2ma$: $3mg-3ma-2mg=2ma$. Thus $mg=5ma$ and $a=\dfrac{g}{5}$.

Using $3mg-T=3m\left(\dfrac{g}{5}\right)$ gives $T=3mg-\dfrac{3mg}{5}=\dfrac{12mg}{5}$.

The two tensions are one vertical and one along the plane. Components of the resultant are $T\cos\alpha$ horizontally and $T+T\sin\alpha$ vertically.

Resultant $=T\sqrt{\cos^2\alpha+(1+\sin\alpha)^2}=T\sqrt{(\tfrac45)^2+(\tfrac85)^2}=\dfrac{4\sqrt5}{5}T$. With $T=\dfrac{12mg}{5}$, this is $\dfrac{48\sqrt5}{25}mg$.

The pulley is smooth, so the tension is the same on both sides of the pulley.

Here $\tan\alpha=\dfrac{5}{12}$, so $\sin\alpha=\dfrac{5}{13}$ and $\cos\alpha=\dfrac{12}{13}$.

For A, $R=2mg\cos\alpha=2mg\cdot\dfrac{12}{13}=\dfrac{24mg}{13}$. Since $\mu=\dfrac{11}{36}$, $F=\mu R=\dfrac{11}{36}\cdot\dfrac{24mg}{13}=\dfrac{22mg}{39}$.

B moves down as A moves up. Taking downward as positive for B: $3mg-T=3ma$.

Taking up the plane as positive for A: $T-2mg\sin\alpha-F=2ma$. Substituting $\sin\alpha=\dfrac{5}{13}$ and $F=\dfrac{22mg}{39}$ gives $T-\dfrac{10mg}{13}-\dfrac{22mg}{39}=2ma$.

From B, $T=3mg-3ma$. Substitute into A equation: $3mg-3ma-\dfrac{10mg}{13}-\dfrac{22mg}{39}=2ma$. The constant part is $\dfrac{5mg}{3}$, so $\dfrac{5mg}{3}=5ma$ and $a=\dfrac{g}{3}$.

While B descends distance $h$, A moves distance $h$ up the plane from rest with acceleration $g/3$. Thus $v^2=2\left(\dfrac{g}{3}\right)h=\dfrac{2gh}{3}$.

After B reaches the ground, the string no longer pulls A. The forces down the plane on A are $2mg\sin\alpha+F=\dfrac{10mg}{13}+\dfrac{22mg}{39}=\dfrac{4mg}{3}$. Since A has mass $2m$, its deceleration is $\dfrac{2g}{3}$.

Let the extra distance be $d$. Using $0=v^2-2\left(\dfrac{2g}{3}\right)d$ with $v^2=\dfrac{2gh}{3}$ gives $d=\dfrac{h}{2}$. Total distance travelled by A is $h+\dfrac{h}{2}=\dfrac{3h}{2}$.

The force of $18$ N is angled away from the plane, so $R+18\sin40^\circ=2g\cos30^\circ$. Hence $R=2g\cos30^\circ-18\sin40^\circ$.

The parcel moves up the plane, so friction acts down the plane. With $\mu=0.3$, $F=0.3R$.

Taking up the plane as positive: $18\cos40^\circ-F-2g\sin30^\circ=2a$.

$18\cos40^\circ-0.3(2g\cos30^\circ-18\sin40^\circ)-2g\sin30^\circ=2a$.

Using $g=9.8$, this gives $a=1.18\ldots\ \mathrm{m\,s^{-2}}$ up the plane.

From A to B, $u=2$, $s=5$ and $a=1.18\ldots$. Use $v^2=u^2+2as$: $v^2=2^2+2(1.18\ldots)(5)$.

$v=3.98\ldots\ \mathrm{m\,s^{-1}}$, about $4.0\ \mathrm{m\,s^{-1}}$.

At C, with the pulling force removed, $R=2g\cos30^\circ$. The maximum friction is $0.3R=0.3(2g\cos30^\circ)=5.09\ldots\ \mathrm{N}$.

The component of weight down the plane is $2g\sin30^\circ=9.8\ \mathrm{N}$. Since $9.8>5.09\ldots$, friction is not large enough to hold P at rest, so P will not remain at rest at C.

Since $\tan\alpha=\dfrac34$, $\sin\alpha=\dfrac35$ and $\cos\alpha=\dfrac45$.

For A, $R=2g\cos\alpha=\dfrac{8g}{5}$. With coefficient $\dfrac14$, friction is $F=\dfrac14R=\dfrac{2g}{5}$.

A moves up the plane. Taking up the plane as positive: $T-2g\sin\alpha-F=2a$.

$T-2g\cdot\dfrac35-\dfrac{2g}{5}=2a$, so $T-\dfrac{8g}{5}=2a$.

B descends. Taking downward as positive: $4g-T=4a$.

From A, $T=2a+\dfrac{8g}{5}$. From B, $T=4g-4a$. Equating gives $6a=\dfrac{12g}{5}$, so $a=\dfrac{2g}{5}$. Then $T=4g-4\left(\dfrac{2g}{5}\right)=\dfrac{12g}{5}$ N.

Before B hits the ground, both particles move distance $h$ from rest with acceleration $\dfrac{2g}{5}$. Thus $v^2=2\left(\dfrac{2g}{5}\right)h=\dfrac{4gh}{5}$.

After B hits the ground, A continues up the plane. The forces down the plane on A are $2g\sin\alpha+F=\dfrac{6g}{5}+\dfrac{2g}{5}=\dfrac{8g}{5}$, so its deceleration is $\dfrac{4g}{5}$.

Let the extra distance be $s$. Then $0=\dfrac{4gh}{5}-2\left(\dfrac{4g}{5}\right)s$, so $s=\dfrac{h}{2}$.

A travels total distance $h+\dfrac{h}{2}=\dfrac{3h}{2}$ before stopping. Since it comes to rest before reaching the pulley, $d>\dfrac{3h}{2}$.

A more realistic model could include friction at the pulley.