(a) When the car moves at constant speed, the driving force is equal to the resisting force.

\[ F=620\text{ N}. \]

Power is given by

\[ P=Fv. \]

So

\[ P=620(25)=15500\text{ W}. \]

\[ P=15.5\text{ kW}. \]

Answer: \(15.5\text{ kW}\).

(b) The new power is

\[ 15.5+12=27.5\text{ kW}=27500\text{ W}. \]

At the instant the power is increased, the speed is still \(25\text{ m s}^{-1}\).

So the new driving force is

\[ F=\frac{P}{v}=\frac{27500}{25}=1100\text{ N}. \]

The resultant force is

\[ 1100-620=480\text{ N}. \]

Using \(F=ma\),

\[ 480=960a. \]

Therefore

\[ a=0.5. \]

Answer: \(0.5\text{ m s}^{-2}\).

(a) Use conservation of momentum for the collision between \(A\) and \(B\).

\[ 2.5(3)=(2.5+3.5)v. \]

\[ 7.5=6v. \]

So

\[ v=1.25\text{ m s}^{-1}. \]

The collision with the wall reduces the speed by \(25\%\), so the rebound speed is \(75\%\) of \(1.25\).

\[ 0.75(1.25)=0.9375. \]

Answer: \(0.938\text{ m s}^{-1}\).

(b) Time for \(C\) to reach the wall:

\[ t_1=\frac{9}{1.25}=7.2\text{ s}. \]

After rebounding, time for \(C\) to be \(9\text{ m}\) from the wall again:

\[ t_2=\frac{9}{0.9375}=9.6\text{ s}. \]

Total time:

\[ t=7.2+9.6=16.8\text{ s}. \]

Answer: \(16.8\text{ s}\).

From rest, \(B\) moves \(0.3\text{ m}\) in \(0.4\text{ s}\).

Using

\[ s=\frac12at^2, \]

\[ 0.3=\frac12a(0.4)^2. \]

So

\[ a=3.75\text{ m s}^{-2}. \]

(a) For particle \(B\), taking down the plane as positive:

\[ 0.8g\sin55^\circ-T=0.8a. \]

\[ 8\sin55^\circ-T=0.8(3.75). \]

\[ T=8\sin55^\circ-3. \]

\[ T=3.553\ldots \]

Answer: \(3.55\text{ N}\).

(b) For particle \(A\), taking motion towards the pulley as positive:

\[ T-\mu R=0.5a. \]

Since \(A\) is on a horizontal plane,

\[ R=0.5g=5\text{ N}. \]

So

\[ 3.553\ldots-5\mu=0.5(3.75). \]

\[ 3.553\ldots-5\mu=1.875. \]

\[ \mu=0.3356\ldots \]

Answer: \(\mu=0.336\).

(c) After \(0.4\text{ s}\), the speed of \(A\) is

\[ v=at=3.75(0.4)=1.5\text{ m s}^{-1}. \]

When the string breaks, the only horizontal force bringing \(A\) to rest is friction.

The work done by friction is equal to the change in kinetic energy:

\[ W=0-\frac12(0.5)(1.5)^2. \]

\[ W=-0.5625\text{ J}. \]

Answer: \(-0.563\text{ J}\).

Since

\[ \tan\theta=\frac43, \]

we have

\[ \sin\theta=\frac45,\qquad \cos\theta=\frac35. \]

(a) Let \(AB=s\).

Initial kinetic energy:

\[ \frac12m(4)^2=8m. \]

Work done against gravity:

\[ mg s\sin\theta=m(10)s\left(\frac45\right)=8ms. \]

Work done against friction:

\[ \mu R s=\frac13mg\cos\theta \cdot s =\frac13m(10)\left(\frac35\right)s=2ms. \]

Total work done against resistance is

\[ 8ms+2ms=10ms. \]

By energy conservation,

\[ 8m=10ms. \]

Therefore

\[ s=0.8. \]

Hence \(AB=0.8\text{ m}\).

(b) On the way up, the deceleration is

\[ 10\sin\theta+\frac13(10)\cos\theta. \]

\[ =10\left(\frac45\right)+\frac13(10)\left(\frac35\right)=8+2=10\text{ m s}^{-2}. \]

Time to reach \(B\):

\[ 0=4-10t. \]

\[ t=0.4\text{ s}. \]

On the way down, the acceleration is

\[ 10\sin\theta-\frac13(10)\cos\theta=8-2=6\text{ m s}^{-2}. \]

Starting from rest and moving \(0.8\text{ m}\):

\[ 0.8=\frac12(6)t^2. \]

\[ t^2=\frac{0.8}{3}. \]

\[ t=0.516\ldots\text{ s}. \]

Total time:

\[ 0.4+0.516\ldots=0.916\ldots \]

Answer: \(0.916\text{ s}\).

Let the time taken by \(P\) to reach the ground be \(T\text{ s}\).

Then the time taken by \(Q\) is \((T-1)\text{ s}\).

For \(P\), using

\[ s=ut+\frac12gt^2, \]

we get

\[ h=\frac12(10)T^2=5T^2. \]

For \(Q\),

\[ h=18(T-1)+\frac12(10)(T-1)^2. \]

So

\[ 5T^2=18(T-1)+5(T-1)^2. \]

Expand:

\[ 5T^2=18T-18+5T^2-10T+5. \]

\[ 5T^2=5T^2+8T-13. \]

Therefore

\[ 8T=13. \]

\[ T=\frac{13}{8}. \]

Hence

\[ h=5\left(\frac{13}{8}\right)^2. \]

\[ h=\frac{845}{64}=13.203125. \]

Answer: \(13.2\text{ m}\).

Since

\[ \sin\theta=\frac5{13}, \]

we have

\[ \cos\theta=\frac{12}{13}. \]

(a) The horizontal component of the tension is

\[ 26\cos\theta=26\left(\frac{12}{13}\right)=24\text{ N}. \]

The vertical component of the tension is

\[ 26\sin\theta=26\left(\frac5{13}\right)=10\text{ N}. \]

Vertically,

\[ R+10=W. \]

So

\[ R=W-10. \]

The frictional force is

\[ 0.25R=0.25(W-10). \]

The mass of the crate is

\[ \frac{W}{10}. \]

Using \(F=ma\) horizontally:

\[ 24-0.25(W-10)=\frac{W}{10}(1.5). \]

\[ 24-0.25W+2.5=0.15W. \]

\[ 26.5=0.4W. \]

\[ W=66.25. \]

Answer: \(W=66.3\text{ N}\).

(b) After the rope is removed, the only horizontal force is friction.

The frictional force is

\[ 0.25W. \]

The mass is \(\frac{W}{10}\), so the deceleration is

\[ a=\frac{0.25W}{W/10}=2.5\text{ m s}^{-2}. \]

Using

\[ v^2=u^2+2as \]

with \(v=0\), \(u=8\), and \(a=-2.5\):

\[ 0=8^2+2(-2.5)s. \]

\[ 0=64-5s. \]

\[ s=12.8. \]

Answer: \(AB=12.8\text{ m}\).

(a) Differentiate \(v\) to find the acceleration:

\[ a=\frac{dv}{dt} =\frac{2k_1}{(4t+1)^{\frac12}}-(2t+1). \]

When \(t=1.25\),

\[ 4t+1=6 \]

and

\[ 2t+1=3.5. \]

The deceleration is \(0.5\text{ m s}^{-2}\), so

\[ a=-0.5. \]

Hence

\[ \frac{2k_1}{\sqrt6}-3.5=-0.5. \]

\[ \frac{2k_1}{\sqrt6}=3. \]

Therefore

\[ k_1=\frac{3\sqrt6}{2}. \]

Answer: \(k_1=\frac{3\sqrt6}{2}\).

(b) Since \(P\) is at instantaneous rest when \(t=3.5\),

\[ v=0 \]

when \(t=3.5\).

So

\[ 0=k_1(15)^{\frac12}-\frac14(8)^2+k_2. \]

Using \(k_1=\frac{3\sqrt6}{2}\),

\[ 0=\frac{3\sqrt6}{2}\sqrt{15}-16+k_2. \]

\[ 0=\frac{9\sqrt{10}}{2}-16+k_2. \]

Therefore

\[ k_2=16-\frac{9\sqrt{10}}{2}. \]

The particle is not at instantaneous rest in \(0\leq t\leq1\), so the distance travelled is

\[ \int_0^1 v\,dt. \]

Thus

\[ \int_0^1\left(k_1(4t+1)^{\frac12}-\frac14(2t+1)^2+k_2\right)\,dt. \]

\[ =\left[\frac{k_1}{6}(4t+1)^{\frac32}-\frac1{24}(2t+1)^3+k_2t\right]_0^1. \]

\[ =\frac{k_1}{6}\left(5^{\frac32}-1\right)-\frac{13}{12}+k_2. \]

Substitute \(k_1=\frac{3\sqrt6}{2}\) and \(k_2=16-\frac{9\sqrt{10}}{2}\):

\[ \text{distance} = \frac{\sqrt6}{4}(5\sqrt5-1)-\frac{13}{12}+16-\frac{9\sqrt{10}}{2}. \]

\[ =6.9205\ldots \]

Answer: \(6.92\text{ m}\).