M2 January 2009 Q7
7. A particle \(P\) of mass \(3m\) is moving in a straight line with speed \(2u\) on a smooth horizontal table. It collides directly with another particle \(Q\) of mass \(2m\) which is moving with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\tfrac{1}{5}(9e + 4)u\). (5)
The speed of \(P\) immediately after the collision is \(\tfrac{1}{2}u\).
(b) Show that \(e = \tfrac{1}{4}\). (4)
The collision between \(P\) and \(Q\) takes place at the point \(A\). After the collision \(Q\) hits a smooth fixed vertical wall which is at right-angles to the direction of motion of \(Q\). The distance from \(A\) to the wall is \(d\).
(c) Show that \(P\) is a distance \(\tfrac{3}{5}d\) from the wall at the instant when \(Q\) hits the wall. (4)
Particle \(Q\) rebounds from the wall and moves so as to collide directly with particle \(P\) at the point \(B\). Given that the coefficient of restitution between \(Q\) and the wall is \(\tfrac{1}{5}\),
(d) find, in terms of \(d\), the distance of the point \(B\) from the wall. (4)
| Scheme | Marks |
|---|---|
| Correct use of NEL | M1* |
| \(y - x = e(2u + u)\) o.e. | A1 |
| CLM \((\rightarrow)\): \(3m(2u) + 2m(-u) = 3m(x) + 2m(y)\ \ (\Rightarrow 4u = 3x + 2y)\) | B1 |
| Hence \(x = y - 3eu\), \(4u = 3(y - 3eu) + 2y\), \((u(9e + 4) = 5y)\) | d*M1 |
| Hence, speed of \(Q = \tfrac{1}{5}(9e + 4)u\) AG | A1 cso |
| (5) |
| Scheme | Marks |
|---|---|
| \(x = y - 3eu = \dfrac{1}{5}(9e + 4)u - 3eu\) | M1# |
| Hence, speed P \(= \dfrac{1}{5}(4 - 6e)u = \dfrac{2u}{5}(2 - 3e)\) o.e. | A1 |
| \(x = \dfrac{1}{2}u = \dfrac{2u}{5}(2 - 3e) \Rightarrow 5u = 8u - 12eu, \Rightarrow 12e = 3\) & solve for \(e\) | d#M1 |
| gives, \(e = \tfrac{3}{12} \Rightarrow e = \tfrac{1}{4}\) AG | A1 |
| (4) |
Or (b)
| Using NEL correctly with given speeds of \(P\) and \(Q\) | M1# |
| \(3eu = \tfrac{1}{5}(9e + 4)u - \tfrac{1}{2}u\) | A1 |
| \(3eu = \tfrac{9}{5}eu + \tfrac{4}{5}u - \tfrac{1}{2}u\), \(3e - \tfrac{9}{5}e = \tfrac{4}{5} - \tfrac{1}{2}\) & solve for \(e\) | d#M1 |
| \(\tfrac{6}{5}e = \tfrac{3}{10} \Rightarrow e = \tfrac{15}{60} \Rightarrow e = \tfrac{1}{4}\). | A1 |
(4)
| Scheme | Marks |
|---|---|
| Time taken by \(Q\) from \(A\) to the wall \(= \dfrac{d}{y} = \left\{\dfrac{4d}{5u}\right\}\) | M1† |
| Distance moved by \(P\) in this time \(= \dfrac{u}{2} \times \dfrac{d}{y}\ \ \left(= \dfrac{u}{2}\left(\dfrac{4d}{5u}\right) = \dfrac{2}{5}d\right)\) | A1 |
| Distance of \(P\) from wall \(= d - x\left(\dfrac{d}{y}\right);\ = d - \tfrac{2}{5}d = \tfrac{3}{5}d\) AG | d†M1; A1 cso |
| (4) |
or (c)
| Ratio speed P:speed Q \(= x : y = \dfrac{1}{2}u : \dfrac{1}{5}\left(\dfrac{9}{4} + 4\right)u = \dfrac{1}{2}u : \dfrac{5}{4}u = 2:5\) | M1† |
| So if \(Q\) moves a distance \(d\), \(P\) will move a distance \(\tfrac{2}{5}d\) | A1 |
| Distance of \(P\) from wall \(= d - \tfrac{2}{5}d;\ = \tfrac{3}{5}d\) AG cso | d†M1; A1 |
(4)
| Scheme | Marks |
|---|---|
| After collision with wall, speed \(Q = \tfrac{1}{5}y = \tfrac{1}{5}\left(\tfrac{5u}{4}\right) = \tfrac{1}{4}u\) their \(y\) | B1ft |
| Time for \(P\), \(T_{AB} = \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u}\), Time for \(Q\), \(T_{WB} = \dfrac{x}{\frac{1}{4}u}\) from their \(y\) | B1ft |
| Hence \(T_{AB} = T_{WB} \Rightarrow \dfrac{\frac{3d}{5} - x}{\frac{1}{2}u} = \dfrac{x}{\frac{1}{4}u}\) | M1 |
| gives, \(2\left(\tfrac{3d}{5} - x\right) = 4x \Rightarrow \tfrac{3d}{5} - x = 2x,\ \ 3x = \tfrac{3d}{5} \Rightarrow x = \tfrac{1}{5}d\) | A1 cao |
| (4) | |
| (17 marks) |
or (d)
| After collision with wall, speed \(Q = \tfrac{1}{5}y = \tfrac{1}{5}\left(\tfrac{5u}{4}\right) = \tfrac{1}{4}u\) their \(y\) | B1ft |
| speed \(P = x = \tfrac{1}{2}u\), speed \(P\): new speed \(Q = \tfrac{1}{2}u : \tfrac{1}{4}u = 2:1\) from their \(y\) | B1ft |
| Distance of \(B\) from wall \(= \dfrac{1}{3} \times \dfrac{3d}{5};\ = \dfrac{d}{5}\) their \(\dfrac{1}{2 + 1}\) | M1; A1 |
(4)
2nd or (d)
| After collision with wall, speed \(Q = \tfrac{1}{5}y = \tfrac{1}{5}\left(\tfrac{5u}{4}\right) = \tfrac{1}{4}u\) their \(y\) | B1ft |
| Combined speed of \(P\) and \(Q = \tfrac{1}{2}u + \tfrac{1}{4}u = \tfrac{3}{4}u\) | |
| Time from wall to 2nd collision \(= \dfrac{\frac{3d}{5}}{\frac{3u}{4}} = \dfrac{3d}{5} \times \dfrac{4}{3u} = \dfrac{4d}{5u}\) from their \(y\) | B1ft |
| Distance of \(B\) from wall = (their speed)x(their time) \(= \dfrac{u}{4} \times \dfrac{4d}{5u};\ = \tfrac{1}{5}d\) | M1; A1 |
(4)