STEP II 2007 question 9 solution

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STEP II 2007 question 9 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 9 solution

(i) The only force on the particle at moment of impact is in direction of the normal, so if the path is parallel to the normal before the collision, it is parallel to the normal after the collision.

Let speed of particle after collision be v, speed of cone be w.

Cons of horiz momentum gives: $Mw - mv \cos \alpha = mu \cos \alpha$

Newton's law of restitution gives: $w \cos \alpha + v = eu$

So $mw \cos^2\alpha + mv \cos \alpha = meu \cos \alpha$

So $(M+m\cos^2\alpha) w = mu(1+e) \cos \alpha$ , so $w =\frac{mu(1+e) \cos \alpha}{M+m\cos^2\alpha}$ .

(ii) Now split the velocity components (post collision) into components v_n, v_t in directions normal to and tangential to the cone surface. Then $v_t = u \cos \alpha$ (particle momentum in this direction being conserved).

Conservation of horiz momentum gives: $Mw - mv_n \cos \alpha = mv_t \sin \alpha$

N's law of restitution gives: $w \cos \alpha +v_n = eu \sin \alpha$ (N.B. resolving initial velocity normal to cone gives approach speed = $u \sin \alpha$ ).

So $mw \cos^2 \alpha + mv_n \cos \alpha = me u \cos \alpha \sin \alpha$ .

So $(M+m\cos^2\alpha)w = m\sin\alpha(eu \cos \alpha + v_t) = m\sin\alpha(1+e)u\cos \alpha$ .

So $\displaystyle w = \frac{mu(1+e)\sin\alpha \cos\alpha}{M+m\cos^2\alpha}$ as required.

For the last part, $\displaystyle \frac{\text{d} w}{\text{d}\alpha} = \frac{(M+mc^2)(c^2-s^2)+2ms^2c^2}{(M+mc^2)^4}$ (where $c=\cos \alpha, s = \sin \alpha$ ).

So $\displaystyle \frac{\text{d} w}{\text{d}\alpha} = 0 \iff (M+mc^2)(c^2-s^2)+2ms^2c^2 = 0$ $\iff M(c^2-s^2)+mc^2(c^2-s^2+2s^2) = 0$ $\iff M(2c^2-1)+mc^2 = 0$ $\iff (2M+m)c^2 = M$

Since $M>0, \cos \alpha>0$ , deduce $\displaystyle \frac{\text{d} w}{\text{d}\alpha} = 0 \iff \cos \alpha = \sqrt{\frac{M}{2M+m}}$ .

So we've shown w has exactly one turning point in $[0,\pi/2]$ . But $w(0)=w(\pi/2) = 0$ , while $w(\pi/4) > 0$ . Thus our turning point must in fact be the unique maximum point for w and so gives the maximum value for w as required.