Challenging +1.2 This is a standard M3 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, conservation of momentum along the line of centres, and Newton's restitution equation. While it involves multiple steps and careful bookkeeping of components, it follows a well-established procedure taught explicitly in M3 with no novel insight required. The given sin α = 0.8 simplifies calculations. Slightly above average difficulty due to the multi-step nature and potential for sign errors, but remains a textbook exercise.
4
\includegraphics[max width=\textwidth, alt={}, center]{a04e6d4e-2437-4761-87ee-43e6771fbbd9-2_332_995_1375_575}
Two uniform smooth spheres \(A\) and \(B\), of equal radius, have masses 4 kg and 3 kg respectively. They are moving on a horizontal surface, and they collide. Immediately before the collision, \(A\) is moving with speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) to the line of centres, where \(\sin \alpha = 0.8\), and \(B\) is moving along the line of centres with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). The coefficient of restitution between the spheres is 0.5 . Find the speed and direction of motion of each sphere after the collision. [0pt]
[10]
For using principle of conservation of momentum in the direction of l.o.c.
Equation complete with not more than one error
A1
\(4a + 3b = 0\)
A1
M1
For using NEL in the direction of l.o.c.
\(0.5(15\cos\alpha + 12) = b - a\)
A1
\([a = -4.5,\; b = 6]\)
M1
For solving for \(a\) and \(b\)
\([\text{Speed} = \sqrt{(-4.5)^2 + 12^2}\), Direction \(\tan^{-1}(12/(-4.50))]\)
M1
For correct method for speed or direction of A
Speed of A is \(12.8\text{ ms}^{-1}\) and direction is \(111°\) anticlockwise from i direction
A1
Direction may be stated in any form, including \(\theta = 69°\) with \(\theta\) clearly and appropriately indicated
Speed of B is \(6\text{ ms}^{-1}\) to the right
A1
[10 marks total] Depends on first three M marks
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 15\sin\alpha \quad (= 12)$ | B1 | |
| $[4(15\cos\alpha) - 3 \times 12 = 4a + 3b]$ | M1 | For using principle of conservation of momentum in the direction of l.o.c. |
| Equation complete with not more than one error | A1 | |
| $4a + 3b = 0$ | A1 | |
| | M1 | For using NEL in the direction of l.o.c. |
| $0.5(15\cos\alpha + 12) = b - a$ | A1 | |
| $[a = -4.5,\; b = 6]$ | M1 | For solving for $a$ and $b$ |
| $[\text{Speed} = \sqrt{(-4.5)^2 + 12^2}$, Direction $\tan^{-1}(12/(-4.50))]$ | M1 | For correct method for speed or direction of A |
| Speed of A is $12.8\text{ ms}^{-1}$ and direction is $111°$ anticlockwise from **i** direction | A1 | Direction may be stated in any form, including $\theta = 69°$ with $\theta$ clearly and appropriately indicated |
| Speed of B is $6\text{ ms}^{-1}$ to the right | A1 | [10 marks total] Depends on first three M marks |
---
4\\
\includegraphics[max width=\textwidth, alt={}, center]{a04e6d4e-2437-4761-87ee-43e6771fbbd9-2_332_995_1375_575}
Two uniform smooth spheres $A$ and $B$, of equal radius, have masses 4 kg and 3 kg respectively. They are moving on a horizontal surface, and they collide. Immediately before the collision, $A$ is moving with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ to the line of centres, where $\sin \alpha = 0.8$, and $B$ is moving along the line of centres with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). The coefficient of restitution between the spheres is 0.5 . Find the speed and direction of motion of each sphere after the collision.\\[0pt]
[10]
\hfill \mbox{\textit{OCR M3 2007 Q4 [10]}}