CAIE A-Level Further Math AS 3.6 Momentum Questions

$$\begin{aligned} \text { PCLM: } m v=-m u \cos 60^{\circ}+2 m u \cos \theta v=-\frac{1}{2} u+2 u \cos \theta=v=\frac{1}{2} u(4 \cos \theta-1) \end{aligned}$$
First line must be seen.

AG

KE of $A=\frac{1}{2} m\left(v^{2}+\left(u \sin 60^{\circ}\right)^{2}\right)$
KE of $A=2 \times \mathrm{KE}$ of B, so
$$\frac{1}{2} m\left(v^{2}+\left(u \sin 60^{\circ}\right)^{2}\right)=2 \times \frac{1}{2} \times 2 m(u \sin \theta)^{2}$$$$\begin{aligned} \left(\frac{1}{2} u(4 \cos \theta-1)\right)^{2}+\frac{3}{4} u^{2}=4 u^{2}(\sin \theta)^{2} 8 \cos ^{2} \theta-2 \cos \theta-3=0 \end{aligned}$$
Use result of (a) and rearrange.
Obtain 3-term quadratic.

$\cos \theta=\frac{3}{4}, \quad-\frac{1}{2}$ but angle is acute, so $\cos \theta=\frac{3}{4}$

NEL: $v=e\left(u \cos 60^{\circ}+u \cos \theta\right)$$$\begin{aligned} v=u, \quad u=e\left(\frac{1}{2} u+\frac{3}{4} u\right) e=\frac{4}{5} \end{aligned}$$

After collision, A has velocity $v_{A}$ towards wall on right and B has component
of velocity towards lower wall of $u \sin \theta$. Same distance and time, so
$v_{A}=u \sin \theta$.
Along line of centres:
$$\begin{array}{ll} \text { PCLM: } m v_{A}+m v_{B}=m u \cos \theta \text { NEL: } v_{A}-v_{B}=e u \cos \theta \end{array}$$
Both, consistent signs, must be $\cos \theta$.

$2 \sin \theta=(1+e) \cos \theta$
Eliminating $v_{A}$ and $v_{B}$ to find an equation in $\theta$. Condone common factor of u.

Note: $v_{A}=\frac{3}{4} u \cos \theta$ and $v_{B}=\frac{1}{4} u \cos \theta$.

$\tan \theta=\frac{3}{4}$
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Final KE $=\frac{1}{2} m\left(v_{A}^{2}+(u \sin \theta)^{2}+v_{B}^{2}\right)$
Both components of velocity of B needed.

Loss $=\frac{1}{2} m u^{2}-\frac{1}{2} m\left(v_{A}^{2}+(u \sin \theta)^{2}+v_{B}^{2}\right)=\frac{1}{2} m u^{2}\left(1-\frac{9}{25}-\frac{9}{25}-\frac{1}{25}\right)$$v_{A}, v_{B}, \theta$ substituted, ft their final KE with both
components of velocity of B included.

Loss $=\frac{3}{25} m u^{2}$

Percentage loss = 24\%
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