CAIE A-Level Further Mathematics 3.6.2 Momentum and Impacts

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
$v_{A}=u \sin \theta$.

B1

Along line of centres:
$$\begin{array}{ll} \text { PCLM: } m v_{A}+m v_{B}=m u \cos \theta \text { NEL: } v_{A}-v_{B}=e u \cos \theta \end{array}$$

M1

Both, consistent signs, must be $\cos \theta$.

$2 \sin \theta=(1+e) \cos \theta$

M1

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 $v_{B}=\frac{1}{4} u \cos \theta$.

$\tan \theta=\frac{3}{4}$

A1

4

Final KE $=\frac{1}{2} m\left(v_{A}^{2}+(u \sin \theta)^{2}+v_{B}^{2}\right)$

B1

Both components of velocity of B needed.

Loss $=\frac{1}{2} m u^{2}-\frac{1}{2} m\left(v_{A}^{2}+(u \sin \theta)^{2}+v_{B}^{2}\right)=\frac{1}{2} m u^{2}\left(1-\frac{9}{25}-\frac{9}{25}-\frac{1}{25}\right)$

M1

$v_{A}, v_{B}, \theta$ substituted, ft their final KE with both
components of velocity of B included.

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

Percentage loss = 24\%

A1

3