4 Two smooth spheres \(P\) and \(Q\), of equal radius, have masses \(m\) and \(3 m\) respectively. They are moving in the same direction in the same straight line on a smooth horizontal table. Sphere \(P\) has speed \(u\) and collides directly with sphere \(Q\) which has speed \(k u\), where \(0 < k < 1\). Sphere \(P\) is brought to rest by the collision. Show that the coefficient of restitution between \(P\) and \(Q\) is \(\frac { 3 k + 1 } { 3 ( 1 - k ) }\). One third of the total kinetic energy of the spheres is lost in the collision. Show that $$k = \frac { 1 } { 3 } ( 2 \sqrt { } 3 - 3 )$$

## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of momentum: $3mv_Q = mu + 3kmu$ | M1 A1 | |
| Newton's law of restitution: $v_Q = e(u - ku)$ | M1 A1 | |
| Eliminate $v_Q$: $e = (3k+1)/3(1-k)$ | M1 A1 | A.G. |
| Relate KE before and after: $\frac{1}{2} \cdot 3mv_Q^2 = \frac{2}{3} \cdot \frac{1}{2}m(u^2 + 3k^2u^2)$ | M1 A1 | |
| Replace $v_Q$ by $\frac{1}{3}(1+3k)u$: $(1+3k)^2 = 2(1+3k^2)$ | | |
| $3k^2 + 6k - 1 = 0$ | M1 A1 | |
| Find root $k$ with $0 < k < 1$: $k = (-6 + \sqrt{48})/6 = \frac{1}{3}(2\sqrt{3}-3)$ | A1 | A.G. Simply substituting given $k$ earns M1 A0 A1 |

**Total: 11 marks**

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4 Two smooth spheres $P$ and $Q$, of equal radius, have masses $m$ and $3 m$ respectively. They are moving in the same direction in the same straight line on a smooth horizontal table. Sphere $P$ has speed $u$ and collides directly with sphere $Q$ which has speed $k u$, where $0 < k < 1$. Sphere $P$ is brought to rest by the collision. Show that the coefficient of restitution between $P$ and $Q$ is $\frac { 3 k + 1 } { 3 ( 1 - k ) }$.

One third of the total kinetic energy of the spheres is lost in the collision. Show that

$$k = \frac { 1 } { 3 } ( 2 \sqrt { } 3 - 3 )$$

\hfill \mbox{\textit{CAIE FP2 2011 Q4 [11]}}