AS June 2023 Q4
4.
Three particles, \(P\), \(Q\) and \(R\), lie at rest on a smooth horizontal plane. The particles are in a straight line with \(Q\) between \(P\) and \(R\), as shown in Figure 1.
Particle \(P\) is projected towards \(Q\) with speed \(u\). At the same time, \(R\) is projected with speed \(\dfrac{1}{2}u\) away from \(Q\), in the direction \(QR\).
Particle \(P\) has mass \(m\) and particle \(Q\) has mass \(2m\).
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
\(\dfrac{u(1+e)}{3}\) (6)
It is given that \(e \gt \dfrac{1}{2}\)
| Scheme | Marks | AO |
|---|---|---|
| \[\begin{array}{cc} u \rightarrow & \leftarrow 0 \\ (P)\ m & (Q)\,2m \\ \longleftarrow & \longrightarrow \\ v_P & v_Q \end{array}\] | ||
| Use of CLM | M1 | 3.4 |
| \(mu = -mv_P + 2mv_Q\) | A1 | 1.1b |
| Use of NEL | M1 | 3.4 |
| \(eu = v_P + v_Q\) | A1 | 1.1b |
| Solve for \(v_Q\) | M1 | 1.1b |
| \(v_Q = \dfrac{u(1+e)}{3}\)* Allow \(\dfrac{u(e+1)}{3}\) | A1* | 2.1 |
| (6) |
Notes
M1: Correct no. of terms with correct pairings, condone sign errors, allow consistently cancelled \(m\)’s or extra \(g\)’s. M0 if they assume the same speeds
A1: Correct equation
M1: Correct no. of terms, condone sign errors. M0 if \(e\) on the wrong side of the equation
A1: Correct equation consistent with their CLM equation
M1: Solve for \(v_Q\)
A1*: Correct given answer correctly obtained
| Scheme | Marks | AO |
|---|---|---|
| \(e \gt \dfrac{1}{2} \Rightarrow \dfrac{u(1+e)}{3} \gt \dfrac{1}{2}u\) Allow this argument reversed. | M1 | 3.1b |
| Hence collision between \(Q\) and \(R\) will occur | A1 | 1.1b |
| (2) |
Notes
M1: Use of \(e \gt \dfrac{1}{2}\) in answer to (a).
Allow M1 for: if \(e = \tfrac{1}{2}\), \(v_Q = \dfrac{1}{2}u\) hence ( if \(e \gt \dfrac{1}{2}\) ), \(v_Q \gt \dfrac{1}{2}u\),
A1: cso
| Scheme | Marks | AO |
|---|---|---|
| \(v_P = \dfrac{u(2e-1)}{3}\) | M1 | 3.1a |
| \(e \gt \dfrac{1}{2} \ \Rightarrow v_P \gt 0\) so \(P\) moves in the opposite direction to its original direction oe e.g. direction (motion) of \(P\) is reversed, opposite direction to \(Q\) A0 for any of: direction changes, \(P\) moves away from \(Q\), goes left, \(P\) will travel in the negative direction, \(P\) moves backwards | A1 | 2.4 |
| ALT: \(e \gt \dfrac{1}{2} \Rightarrow v_Q \gt \dfrac{1}{2}u\) oe Since \(v_P = 2v_Q - u,\ v_P \gt 0\) | M1 | |
| so \(P\) moves in the opposite direction to its original direction oe e.g. direction (motion) of \(P\) is reversed, opposite direction to \(Q\) | A1 | |
| (2) |
Notes
M1: Finds \(v_P\) in terms of \(u\) and \(e\) only using one of their equations
A1: cso
| Scheme | Marks | AO |
|---|---|---|
| Correct expression using their \(v_P\) | M1 | 3.1a |
| \(\dfrac{1}{2}mu^2 - \dfrac{1}{2}m\left[\dfrac{u(2e-1)}{3}\right]^2 - \dfrac{1}{2}2m\left[\dfrac{u(e+1)}{3}\right]^2\) | A1 | 1.1b |
| \(= \dfrac{1}{3}mu^2 - \dfrac{1}{3}mu^2e^2 = \dfrac{1}{3}mu^2(1 - e^2)\) | A1 | 1.1b |
| (3) |
Notes
M1: Correct no. of terms, dimensionally correct. Allow if \(m\) and \(2m\) are swapped or \(m\) used for mass of \(Q\) but otherwise correct.
Allow an expression for the KE gain.
A1: Correct unsimplified expression
A1: Any correct simplified two term expression, not necessarily factorised.
| Scheme | Marks | AO |
|---|---|---|
| The answer for part (d) should equal 0 when \(e = 1\). | B1 | 2.4 |
| (1) | ||
| (14 marks) |
Notes
B1: Correct explanation. This can be scored after an incorrect answer to (d).
Need to say that when \(e = 1\), KE change = 0
Not enough to say KE Before = KE After