M1 June 2013 (R) Q4
4. At time \(t = 0\), two balls \(A\) and \(B\) are projected vertically upwards. The ball \(A\) is projected vertically upwards with speed 2 m s\(^{-1}\) from a point 50 m above the horizontal ground. The ball \(B\) is projected vertically upwards from the ground with speed 20 m s\(^{-1}\). At time \(t = T\) seconds, the two balls are at the same vertical height, \(h\) metres, above the ground. The balls are modelled as particles moving freely under gravity. Find
| Scheme | Marks |
|---|---|
| Use of \(s = ut + \dfrac{1}{2}at^2\) | M1 |
| \(-2t + \dfrac{1}{2}gt^2\) (+ or – 50) | A1 |
| \(20t - \dfrac{1}{2}gt^2\) (+ or – 50) | A1 |
| \(50 = -2T + \dfrac{1}{2}gT^2 + 20T - \dfrac{1}{2}gT^2 = 18T\) | M1 |
| \(T = \dfrac{50}{18} = 2.777\ldots = 2.8\) or better | A1 |
| (5) |
Notes
First M1 for use of \(s = ut + 1/2at^2\) (or use of 2 suvat formulae AND eliminating \(v\), to give an equation in \(s\) and \(t\)). N.B. M0 if they use \(s = 50\) or \(u = 0\) or \(v = 0\))
First A1 with \(u = 2\) and \(a = -g\) or \(-9.8\) to obtain a distance, possibly with 50 added or subtracted. (2 and 4.9 must have opposite signs)
Second A1 with \(u = 20\) and \(a = -g\) or \(-9.8\) to obtain a distance, possibly with 50 added or subtracted. (2 and 4.9 must have opposite signs)
Second M1 dependent on first M1 for a correct equation obtained correctly in \(T\) only.
Third A1 for 25/9 oe, 2.8 or better
| Scheme | Marks |
|---|---|
| \(h = 20 \times T - 4.9 \times T^2 = 17.74\ldots \approx 17.7\) (18 to 2 s.f.) (use of 2.8 gives 17.584) | M1A1 |
| (2) | |
| (7 marks) |
Notes
First M1 for substituting their \(T\) value (allow –ve changed to +ve but A mark is then unavailable) into an appropriate equation
First A1 for 17.7 or 18 (m). (A0 if they then add 50)