Standard +0.8 This is an oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, application of both momentum conservation and Newton's restitution law, then vector recombination. It involves multiple steps with careful component work and is more demanding than standard direct impact questions, but follows a well-established method taught in M3.
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\includegraphics[max width=\textwidth, alt={}, center]{af1f9f1b-f6c0-4044-9864-5b9ce309d3fa-02_283_711_1754_722}
Two uniform smooth spheres \(A\) and \(B\), of equal radius, have masses 5 kg and 2 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision, \(A\) has speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is moving perpendicular to the line of centres, and \(B\) has speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along the line of centres (see diagram). The coefficient of restitution is 0.75 . Find the speed and direction of motion of each sphere immediately after the collision.
[25A² = 1.6² or ½ (0.2)1.6² + 3.5x0.392²/(2x0.7) + 0.2gA = 3.5x(0.392 + A)²/(2x0.7)]
M1
For using A²n² = v²max or Energy at lowest point = energy at equilibrium point (4 terms needed including 2 EE terms)
Amplitude is 0.32m
A1ft [5]
(iii)
Answer
Marks
Guidance
[x = 0.32sin2]
M1
For using x = Asin nt or Acos(\(\pi/2 - nt\))
x = 0.291
A1
[v = 0.32x5cos2° or v² = 25(0.32² - 0.291²) or 0.256 + 0.38416 + 0.2g(0.291) = ½ 0.2v² + 2.5(0.683)²]
M1
For using v = A\ncos nt or v² = n²(A² - x²) or Energy at equilibrium point = energy at x = 0.291
v² = 0.443
A1
May be implied
v = -0.666 (or 0.666 upwards)
A1 [5]
**(i)**
| | M1 | For using T = mg and T = λ e/L |
|---|---|---|
| 3.5e0/7 = 0.2g [e = 0.392] | A1 | |
|---|---|---|
| Position is 1.092m below O. | A1 [3] AG |
**(ii)**
| | M1 | For using Newton's second law |
|---|---|---|
| 0.2g - 3.5(0.392 + x)/0.7 = 0.2a | A1ft | ft incorrect e |
|---|---|---|
| a = -25x | A1ft | ft incorrect e |
|---|---|---|
| [25A² = 1.6² or ½ (0.2)1.6² + 3.5x0.392²/(2x0.7) + 0.2gA = 3.5x(0.392 + A)²/(2x0.7)] | M1 | For using A²n² = v²max or Energy at lowest point = energy at equilibrium point (4 terms needed including 2 EE terms) |
|---|---|---|
| Amplitude is 0.32m | A1ft [5] |
**(iii)**
| [x = 0.32sin2] | M1 | For using x = Asin nt or Acos($\pi/2 - nt$) |
|---|---|---|
| x = 0.291 | A1 | |
|---|---|---|
| [v = 0.32x5cos2° or v² = 25(0.32² - 0.291²) or 0.256 + 0.38416 + 0.2g(0.291) = ½ 0.2v² + 2.5(0.683)²] | M1 | For using v = A\ncos nt or v² = n²(A² - x²) or Energy at equilibrium point = energy at x = 0.291 |
|---|---|---|
| v² = 0.443 | A1 | May be implied |
|---|---|---|
| v = -0.666 (or 0.666 upwards) | A1 [5] |
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\includegraphics[max width=\textwidth, alt={}, center]{af1f9f1b-f6c0-4044-9864-5b9ce309d3fa-02_283_711_1754_722}
Two uniform smooth spheres $A$ and $B$, of equal radius, have masses 5 kg and 2 kg respectively. They are moving on a horizontal surface when they collide. Immediately before the collision, $A$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is moving perpendicular to the line of centres, and $B$ has speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the line of centres (see diagram). The coefficient of restitution is 0.75 . Find the speed and direction of motion of each sphere immediately after the collision.
\hfill \mbox{\textit{OCR M3 Q4 [11]}}