| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 13 |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Standard +0.3 This is a standard FM1 collision question with straightforward application of conservation of momentum, coefficient of restitution formula, and impulse calculation. Part (i) requires basic reasoning about direction, parts (ii-iii) are routine calculations, part (iv) is recall of e<1 meaning, and parts (v-vi) apply impulse formula with Newton's third law. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average for FM1. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Initially, total momentum is in direction of motion of B, so final total momentum is also in this direction; hence both cannot be moving in the opposite direction, which would be the case if A's direction of motion was not reversed | M1, E1 [2] | For considering total momentum and realising relevance of 2.1 × 3 being greater than 1.3 × 2, oe; Full explanation required |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_A = -4\) | B1 | oe, eg use of opposite signs for velocities of A before and after impact |
| \(1.4 \times 2 + 2.1 \times (-3) = 1.4 \times (-4) + 2.1v_B\) | M1 | Use of conservation of momentum with 4 terms; allow 1 sign error for M mark |
| \(v_B = 1\) in opposite direction to B's original motion | A1, A1 [4] | Completely correct equation; Clear indication of direction required |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_B - v_A = -e(2-(-3))\) \(e = 1\) | M1, A1 [2] | With their numerical \(v_A\) and \(v_B\) |
| Answer | Marks | Guidance |
|---|---|---|
| No KE is lost in the collision (as it is perfectly elastic) | EIft [1] | FT conclusion so long as their \(e \in [0, 1]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\pm(2.1 \times 1 - 2.1 \times (-3))\) or \(\pm(1.4 \times (-4) - 1.4 \times 2)\) (Impulse is) 8.4 (N s) So it will be damaged | M1, A1, EIft [3] | Change of momentum for either particle; Impulse is 8.4 (N s) |
| Answer | Marks | Guidance |
|---|---|---|
| They have the same 'damage threshold' and equal and opposite impulses act on the particles in the collision | E1 [1] | Allow quote of 'Newton's Third Law' for second aspect |
## (i)
Initially, total momentum is in direction of motion of B, so final total momentum is also in this direction; hence both cannot be moving in the opposite direction, which would be the case if A's direction of motion was not reversed | M1, E1 [2] | For considering total momentum and realising relevance of 2.1 × 3 being greater than 1.3 × 2, oe; Full explanation required | Positive/negative directions should be defined by candidates or may be inferred from a momentum equation or other evidence
## (ii)
$v_A = -4$ | B1 | oe, eg use of opposite signs for velocities of A before and after impact
$1.4 \times 2 + 2.1 \times (-3) = 1.4 \times (-4) + 2.1v_B$ | M1 | Use of conservation of momentum with 4 terms; allow 1 sign error for M mark
$v_B = 1$ in opposite direction to B's original motion | A1, A1 [4] | Completely correct equation; Clear indication of direction required
## (iii)
$v_B - v_A = -e(2-(-3))$ $e = 1$ | M1, A1 [2] | With their numerical $v_A$ and $v_B$
## (iv)
No KE is lost in the collision (as it is perfectly elastic) | EIft [1] | FT conclusion so long as their $e \in [0, 1]$
## (v)
$\pm(2.1 \times 1 - 2.1 \times (-3))$ or $\pm(1.4 \times (-4) - 1.4 \times 2)$ (Impulse is) 8.4 (N s) So it will be damaged | M1, A1, EIft [3] | Change of momentum for either particle; Impulse is 8.4 (N s)
## (vi)
They have the same 'damage threshold' and equal and opposite impulses act on the particles in the collision | E1 [1] | Allow quote of 'Newton's Third Law' for second aspect
---5 Two discs, $A$ and $B$, have masses 1.4 kg and 2.1 kg respectively. They are sliding towards each other in the same straight line across a large sheet of horizontal ice. Immediately before the collision $A$ has speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ has speed $3 \mathrm {~ms} ^ { - 1 }$. Immediately after the collision $A$ 's speed is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Explain why it is impossible for $A$ to be travelling in the same direction after the collision as it was before the collision.\\
(ii) Find the velocity of $B$ immediately after the collision.\\
(iii) Calculate the coefficient of restitution between $A$ and $B$.\\
(iv) State what your answer to part (iii) means about the kinetic energy of the system.
The discs are made from the same material. The discs will be damaged if subjected to an impulse of magnitude greater than 6.5 Ns .\\
(v) Determine whether $B$ will be damaged as a result of the collision.\\
(vi) Explain why $A$ will be damaged if, and only if, $B$ is damaged.
\hfill \mbox{\textit{OCR FM1 AS 2017 Q5 [13]}}