AS June 2023 Q3
3. A stone of mass 0.5 kg is projected vertically upwards with a speed \(U\ \text{m s}^{-1}\) from a point \(A\). The point \(A\) is 2.5 m above horizontal ground.
The speed of the stone as it hits the ground is \(25\ \text{m s}^{-1}\)
The motion of the stone from the instant it is projected from \(A\) until the instant it hits the ground is modelled as that of a particle moving freely under gravity.
In reality, the stone will be subject to air resistance as it moves from \(A\) to the ground.
The ground is soft and the stone sinks a vertical distance \(d\) cm into the ground. The resistive force exerted on the stone by the ground is modelled as a constant force of magnitude 2000 N and the stone is modelled as a particle.
| Scheme | Marks | AO |
|---|---|---|
| Attempt at use of conservation of energy principle | M1 | 3.4 |
| \(\dfrac{1}{2}m \times 25^2 - \dfrac{1}{2}mU^2 = mg \times 2.5\) | A1 A1 | 1.1b 1.1b |
| \(U = 24\) | A1 | 1.1b |
| ALT 1: \(\dfrac{1}{2}m \times 25^2 = mgh\) \(0 = U^2 - 2g\left(\tfrac{625}{2g} - 2.5\right)\) \(U = 24\) ALT 2: \(\dfrac{1}{2}m \times U^2 = mgd\) \(25^2 = 2g\left(\tfrac{U^2}{2g} + 2.5\right)\) M1A2 for a complete method (-1 eeoo) \(U = 24\) ALT 3: \(\dfrac{1}{2}m \times U^2 = mgd\) \(\dfrac{1}{2}m \times 25^2 = mg\left(\tfrac{U^2}{2g} + 2.5\right)\) \(U = 24\) | ||
| (4) |
Notes
M1: Correct no. of terms (two KE and one PE), dimensionally correct equation, condone sign errors
\(m\) does not need to be substituted and allow cancelled \(m\)’s.
N.B. M0 if clearly using \(v^2 = u^2 + 2as\) for the whole motion.
M0 if they use \(0.5g \times (2.5 + 0.01d)\) since extra term
A1: Correct equation in \(U\) only with at most one error
A1: Correct equation in \(U\) only
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| The value of \(U\) would be larger | B1 | 3.5a |
| (1) |
Notes
B1: cao
| Scheme | Marks | AO |
|---|---|---|
| WD against resistance \(= 2000 \times 0.01d\) | M1 | 3.4 |
| Use of work-energy principle | M1 | 3.1b |
| \(2000 \times 0.01d = 0.5g \times 0.01d + \dfrac{1}{2} \times 0.5 \times 25^2\) | A1 A1 | 1.1b 1.1b |
| \(d = 7.83\) (3 sf) N.B. If PE term is omitted, \(d = 7.8152\) and scores max M1M0A0A0A0 | A1 | 1.1b |
| (5) | ||
| (10 marks) |
Notes
M1: Use of work = force x distance (allow if 0.01 is omitted)
M1: Correct no. of terms (one KE, one PE, one Work Done), dimensionally correct equation
N.B. M0 if not using work-energy principle
A1: Correct equation in \(d\) only with at most one error (omission of 0.01 twice is one error)
A1: Correct equation in \(d\) only
A1: cao