Question 1
Topic – ALV: 4.5
A particle of mass 1.6 kg is dropped from a height of 9 m above horizontal ground. The speed of the particle at the instant before hitting the ground is $12 \ ms^{-1}$.
Find the work done against air resistance.
▶️Answer/Explanation
Solution :-
\begin{align*}
12^2 &= 2 \times 9 \times a \quad \text{OR} \quad a = 8 \\
1.6g – R &= 1.6a \\
\text{WD} &= 3.2 \times 9 = 28.8 \text{ J} \\
(\text{KE}) &= \frac{1}{2} \times 1.6 \times 12^2 \quad \text{OR} \quad 115.2 \\
(\text{Loss of PE}) &= \pm 1.6g \times 9 \quad \text{OR} \quad \pm 144 \\
\text{WD} &= 28.8 \text{ J}
\end{align*}
Question 2
Topic – ALV: 4.3
Two particles A and B, of masses 3.2 kg and 2.4 kg respectively, lie on a smooth horizontal table.
A moves towards B with a speed of $v \ ms^{-1}$ and collides with B, which is moving towards A with a speed of $6 \ ms^{-1}$. In the collision the two particles come to rest.
(a) Find the value of $v$.
(b) Find the loss of kinetic energy of the system due to the collision.
▶️Answer/Explanation
Solution :-
(a)\begin{align*}
\pm[3.2v + 2.4 \times (-6)] &= 0 \\
v &= 4.5
\end{align*}
(b)\begin{align*}
\text{KE} &= \pm \frac{1}{2} \times 3.2 \times (4.5)^2 \quad \text{OR} \quad \pm \frac{1}{2} \times 2.4 \times 6^2 \\
\text{KE}_\text{loss} &= 75.6 \text{ J}
\end{align*}
Question 3
Topic – ALV: 4.1
Coplanar forces of magnitudes 30N, 15N, 33N and PN act at a point in the directions shown in the diagram, where $\tan~\alpha=\frac{4}{3}$. The system is in equilibrium.
(a) Show that
\[\left(\frac{14.4}{30-P}\right)^{2}+\left(\frac{28.8}{P+30}\right)^{2}=1.\]
(b) Verify that $P=6$ satisfies this equation and find the value of $\theta$.
▶️Answer/Explanation
Solution :-
(a) \begin{align*}
(33+15)\times \frac{3}{5} &= P\cos\theta + 30\cos\theta \\
\text{OR} \quad (33+15)\cos\left(\tan^{-1}\frac{4}{3}\right) &= P\cos\theta + 30\cos\theta \\
\text{OR} \quad 19.8 + 9 &= P\cos\theta + 30\cos\theta \\
15\times\frac{4}{5} + 30\sin\theta &= 33\times\frac{4}{5} + P\sin\theta \\
\text{OR} \quad 15\sin\left(\tan^{-1}\frac{4}{3}\right) + 30\sin\theta &= 33\sin\left(\tan^{-1}\frac{4}{3}\right) + P\sin\theta \\
\text{OR} \quad 12 + 30\sin\theta &= 26.4 + P\sin\theta \\
\left[\text{Use } \cos^2\theta + \sin^2\theta = 1 \text{ with } \cos\theta = \frac{28.8}{P+30} \text{ and } \sin\theta = \frac{14.4}{30-P} \text{ to get }\right] \\
\left(\frac{14.4}{30-P}\right)^2 + \left(\frac{28.8}{P+30}\right)^2 &= 1
\end{align*}
(b) $$\text{Sub P=6 into}$$
\[\left(\frac{14.4}{30-P}\right)^{2}+\left(\frac{28.8}{P+30}\right)^{2}\]
$$\text{to get}$$
\[\left[\left(\frac{14.4}{24}\right)^{2}+\left(\frac{28.8}{36}\right)^{2}\right]=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=0.36+0.64=1\]
$$\theta=36.9$$
Question 4
Topic – ALV: 4.5
An athlete of mass 84 kg is running along a straight road.
(a) Initially the road is horizontal and he runs at a constant speed of $3 \ ms^{-1}$. The athlete produces a constant power of 60 W.
Find the resistive force which acts on the athlete.
(b) The athlete then runs up a 150m section of the road which is inclined at $0.8^{\circ}$ to the horizontal. The speed of the athlete at the start of this section of road is $3 \ ms^{-1}$ and he now produces a constant driving force of 24 N. The total resistive force which acts on the athlete along this section of road has constant magnitude 13 N.
Use an energy method to find the speed of the athlete at the end of the 150 m section of road.
▶️Answer/Explanation
Solution :-
(a) $$\text{Resistive force} = DF = \frac{60}{3} = 20 \text{ N}$$
(b) \begin{align}
\text{PE} &= \pm 84g \times 150 \sin(0.8) \\
\text{KE change} &= \pm \left( \frac{1}{2} \times 84v^2 – \frac{1}{2} \times 84 \times 3^2 \right) \\
\text{Work Done} &= \pm (24 \times 150 – 13 \times 150) \\
84g \times 150 \sin(0.8) + \frac{1}{2} \times 84v^2 – \frac{1}{2} \times 84 \times 3^2 &= 24 \times 150 – 13 \times 150 \\
[1759.23… + 42v^2 – 378 &= 3600 – 1950 \rightarrow 42v^2 = 268.765…] \\
[v] &= 2.53 \text{ ms}^{-1} \\
\end{align}
Question 5
Topic – ALV: 4.1
A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of $35^{\circ}$ to the horizontal. The particle is kept in equilibrium by a horizontal force of magnitude $P$ N acting in a vertical plane containing a line of greatest slope (see diagram). The coefficient of friction between the particle and plane is 0.4.
Find the least possible value of $P$.
▶️Answer/Explanation
Solution :-
$\text{Attempt at resolving parallel or perpendicular to the plane.}$
$R = P\sin35 + 0.6g\cos35$
$R = (0.573…)P + 4.914…$
$F + P\cos35 = 0.6g\sin35$
$F + (0.819…)P = 3.441…$
$\text{Use of } F = 0.4R$
$\text{Solve for P}$
$P=1.41$
Question 6
Topic – ALV: 4.2
A particle P starts at rest and moves in a straight line from a point O. At time $t$ s after leaving O, the velocity of P, $v \ ms^{-1}$, is given by $v=bt+ct^{\frac{3}{2}}$, where $b$ and $c$ are constants. P has velocity $8 \ ms^{-1}$ when $t=4$ and has velocity $13.5 \ ms^{-1}$ when $t=9$
(a) Show that $b=3$ and $c=-0.5$.
(b) Find the acceleration of P when $t=1$.
(c) Find the positive value of $t$ when P is at instantaneous rest and find the distance of P from O at this instant.
(d) Find the speed of P at the instant it returns to O.
▶️Answer/Explanation
Solution :-
(6c) Equate v to 0 and attempt to solve for t
t=36 ONLY
Attempt to integrate
$s = \frac{3}{2}t^{2}-0.5\times\frac{2}{5}\times t^{\frac{5}{2}}(+D)=\frac{3}{2}t^{2}-\frac{1}{5}t^{\frac{5}{2}}(+D)$
OR
$s = \frac{b}{2}t^{2}+c\times\frac{2}{5}\times t^{\frac{5}{2}}(+D)$
Sub t=36 (or use limits 36 and 0) to get distance = 388.8 m ONLY
(6d) $ \frac{3}{2}t^{2}-0.5\times\frac{2}{5}\times t^{\frac{5}{2}}=0$
t=56.25
Speed = $42.2 ms^{-1}$ ONLY
Question 7
Topic – ALV: 4.4
Two particles P and Q, of masses 2 kg and 0.25 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. Particle P is on an inclined plane at an angle of 30^{\circ} to the horizontal. Particle Q hangs below the pulley. Three points A, B and C lie on a line of greatest slope of the plane with AB = 0.8 m and BC = 1.2 m (see diagram).
Particle P is released from rest at A with the string taut and slides down the plane. During the motion of P from A to C, Q does not reach the pulley. The part of the plane from A to B is rough, with coefficient of friction 0.3 between the plane and P. The part of the plane from B to C is smooth.
(a) (i) Find the acceleration of P between A and B.
(ii) Hence, find the speed of P at C.
(b) Find the time taken for P to travel from A to C.
▶️Answer/Explanation
Solution :-
7(a) (i) $ \text{Particle P:} 2g\sin(30) – F – T = 2a$
$10 – F – T = 2a$
$ \text{Particle Q:} T – 0.25g = 0.25a $
$ \text{System:} 2g\sin(30) – F – 0.25g = (2 + 0.25)a $
$ F = 0.3R = 0.3 \times 2g \cos(30)$
$= 3\sqrt{3} = 5.1961…$
$ \text{Acceleration from A to B} = 1.02 \text{ ms}^{-2} $
(a)(ii) $ \text{Use of suvat from A to B to get an equation in } v^{2} \text{ OR } v$
$ v^{2}=\frac{80-32\sqrt{3}}{15}=1.64\text{OR} v=1.28 $
$ \text{Find the acceleration from B to C:} 2g\sin(30) – 0.25g = (2+0.25)a $
$ v^{2}=(1.28)^{2}+2\times\frac{10}{3}\times1.2 $
$ \text{Velocity} = 3.1(0) \text{ ms}^{-1} $
$\text{Alternative Method for Question 7(a)(ii): using suvat in first stage and energy in second stage}$
$ \text{Use of suvat from A to B to get an equation in } v^{2} \text{ or } v. $
$ v^{2}=\frac{80-32\sqrt{3}}{15}]=1.64 \text{OR} v=1.28 $
$ \text{Change in PE} = \pm(2g \times 1.2 \sin{30} – 0.25g \times 1.2) $
$ \text{OR Change in KE} = \pm \left[ \frac{1}{2}(2+0.25)v^{2} – \frac{1}{2}(2+0.25)(their~1.28)^{2} \right] $
$ \frac{1}{2}(2+0.25)v^{2} – \frac{1}{2}(2+0.25)(their~1.28)^{2} = 2g \times 1.2 \sin{30} – 0.25g \times 1.2 $
$ \text{Velocity} = 3.1(0) \text{ ms}^{-1} $
7(b) $0.8 = 0 + \frac{1}{2} \times (their \ positive \ answer \ to \ (a)(i)) \times t_{1}^{2}$ and solve for $t_{1}$
OR $0.8 = \frac{1}{2}(0 + their \ positive \ 1.28 \ from \ (a)(ii)) \times t_{1}$ and solve for $t_{1}$
OR
$(their \ positive \ 1.28 \ from \ (a)(ii)) = (their \ positive \ answer \ to \ (a)(i)) \times t_{1}$ and solve for $t_{1}$
$1.2 = (their \ 1.28 \ from \ (a)(ii)) \times t_{2} + \frac{1}{2} \times (their) \times \frac{10}{3} \times t_{2}^{2}$ and solve for $t_{2}$
OR
$1.2 = \frac{1}{2}(their \ 1.28 \ from \ (a)(ii)) + (their \ answer \ to \ (a)(ii)) \times t_{2}$ and solve for $t_{2}$
OR
$(their \ answer \ to \ (a)(ii)) = (their \ 1.28 \ from \ (a)(ii)) + (their) \times \frac{10}{3} \times t_{2}$ and solve for $t_{2}$
$t_{1}=1.25 \ or \ t_{2}=0.547$
Total time = 1.8s