| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision on inclined plane |
| Difficulty | Standard +0.3 This is a standard M1 collision problem requiring conservation of momentum applied twice, with straightforward kinematics on an inclined plane. The multi-part structure and bookkeeping of velocities adds modest complexity, but all techniques are routine and the problem follows a predictable template with no novel insights required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity of P just before collision: \(v_P = 3 - 2.5(0.4) = 2\) m s\(^{-1}\) up plane | M1 A1 | |
| Velocity of Q just before collision: \(v_Q = 0 + 2.5(0.4) = 1\) m s\(^{-1}\) down plane | M1 A1 | |
| Taking up as positive: \(0.5(2) + 0.2(-1) = 0.5v_P' + 0.2(3.2)\) | M1 | CLM |
| \(1 - 0.2 = 0.5v_P' + 0.64\) | ||
| \(0.5v_P' = 0.16\), \(v_P' = 0.32\) m s\(^{-1}\) up the plane | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity of Q, 0.6 s after first collision: \(v_Q = 3.2 - 2.5(0.6) = 1.7\) m s\(^{-1}\) up plane | M1 A1 | |
| Velocity of R just before: \(0 + 2.5(1) = 2.5\) m s\(^{-1}\) down plane (R released at t=0, hits at t=1) | M1 A1 | |
| \((0.2)(1.7) + (0.3)(-2.5) = (0.5)v\) | M1 | CLM |
| \(0.34 - 0.75 = 0.5v\) | ||
| \(v = -0.82\) m s\(^{-1}\), i.e. \(0.82\) m s\(^{-1}\) down the plane | A1 |
# Question 5:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity of P just before collision: $v_P = 3 - 2.5(0.4) = 2$ m s$^{-1}$ up plane | M1 A1 | |
| Velocity of Q just before collision: $v_Q = 0 + 2.5(0.4) = 1$ m s$^{-1}$ down plane | M1 A1 | |
| Taking up as positive: $0.5(2) + 0.2(-1) = 0.5v_P' + 0.2(3.2)$ | M1 | CLM |
| $1 - 0.2 = 0.5v_P' + 0.64$ | | |
| $0.5v_P' = 0.16$, $v_P' = 0.32$ m s$^{-1}$ up the plane | A1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity of Q, 0.6 s after first collision: $v_Q = 3.2 - 2.5(0.6) = 1.7$ m s$^{-1}$ up plane | M1 A1 | |
| Velocity of R just before: $0 + 2.5(1) = 2.5$ m s$^{-1}$ down plane (R released at t=0, hits at t=1) | M1 A1 | |
| $(0.2)(1.7) + (0.3)(-2.5) = (0.5)v$ | M1 | CLM |
| $0.34 - 0.75 = 0.5v$ | | |
| $v = -0.82$ m s$^{-1}$, i.e. $0.82$ m s$^{-1}$ down the plane | A1 | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{ce4c43e6-da4f-4c02-ab0f-01a21717949c-3_362_1065_258_539}
Three particles $P , Q$ and $R$ lie on a line of greatest slope of a smooth inclined plane. $P$ has mass 0.5 kg and initially is at the foot of the plane. $R$ has mass 0.3 kg and initially is at the top of the plane. $Q$ has mass 0.2 kg and is between $P$ and $R$ (see diagram). $P$ is projected up the line of greatest slope with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the instant when $Q$ and $R$ are released from rest. Each particle has an acceleration of $2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ down the plane.\\
(i) $P$ and $Q$ collide 0.4 s after being set in motion. Immediately after the collision $Q$ moves up the plane with speed $3.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the speed and direction of motion of $P$ immediately after the collision.\\
(ii) 0.6 s after its collision with $P , Q$ collides with $R$ and the two particles coalesce. Find the speed and direction of motion of the combined particle immediately after the collision
\hfill \mbox{\textit{OCR M1 2011 Q5 [10]}}