Challenging +1.2 This is a multi-stage collision problem requiring systematic application of conservation of momentum and restitution equations across three events (A-B collision, B-barrier collision, second A-B collision). While it involves several steps and careful bookkeeping of velocities, the techniques are standard for Further Maths mechanics: each collision uses the same two equations (momentum conservation and Newton's experimental law). The final distance calculation requires basic kinematics. The problem is methodical rather than conceptually challenging, making it moderately above average difficulty.
\includegraphics{figure_3}
Two perfectly elastic small smooth spheres \(A\) and \(B\) have masses \(3m\) and \(m\) respectively. They lie at rest on a smooth horizontal plane with \(B\) at a distance \(a\) from a smooth vertical barrier. The line of centres of the spheres is perpendicular to the barrier, and \(B\) is between \(A\) and the barrier (see diagram). Sphere \(A\) is projected towards sphere \(B\) with speed \(u\) and, after the collision between the spheres, \(B\) hits the barrier. The coefficient of restitution between \(B\) and the barrier is \(\frac{1}{4}\). Find the speeds of \(A\) and \(B\) immediately after they first collide, and the distance from the barrier of the point where they collide for the second time. [9]
Find rebound speed of B after collision with barrier: w = ½ v [= 3u/4] M1
B B
EITHER: Find time for B to colln. at d from barrier: t1 = a / v [= 2a/3u] and
B
t2 = d / w [= 8a/15u] B1
B
Find time for A to same collision: t1 + t2 = (a – d) / v B1
A
Equate times and solve for d: 2(a – d) = 2a/3 + 4d/3, d = 2a/5 M1 A1
OR: Find dist. A moves in time t1: s = v × (a/v ) [= a/3] (B1)
A A B
Find t2 from both A and B: t2 = (a – s – d) / v , t2 = d / w (B1)
A A B
Equate times and solve for d: 2(2a/3 – d)/u = 4d/3u, d = 2a/5 (M1 A1)
MR: Taking v – v = – ½ u gives: v = 5u/8, v = 9u/8, w = 9u/16
A B A B B
t1 = 8a/9u , t2 = 64a/171u, d = 4a/19
or taking v – v = – e u gives: v = (3 – e)u/4, v = 3(e + 1)u/4
A B A B
w = 3(e + 1)u/8, t1 = 4a/3(e + 1)u
B
t2 = 32ea/3(e + 1)(e + 9)u
Answer
Marks
d = 4ea/(e + 9) (max 8)
5
4
[9]
Question 3:
3 | Use conservation of momentum: 3mv + mv = 3mu B1
A B
Use Newton’s law of restitution: v – v = – u B1
A B
Solve for v and v : v = ½ u and v = 3u/2 M1 A1
A B A B
Find rebound speed of B after collision with barrier: w = ½ v [= 3u/4] M1
B B
EITHER: Find time for B to colln. at d from barrier: t1 = a / v [= 2a/3u] and
B
t2 = d / w [= 8a/15u] B1
B
Find time for A to same collision: t1 + t2 = (a – d) / v B1
A
Equate times and solve for d: 2(a – d) = 2a/3 + 4d/3, d = 2a/5 M1 A1
OR: Find dist. A moves in time t1: s = v × (a/v ) [= a/3] (B1)
A A B
Find t2 from both A and B: t2 = (a – s – d) / v , t2 = d / w (B1)
A A B
Equate times and solve for d: 2(2a/3 – d)/u = 4d/3u, d = 2a/5 (M1 A1)
MR: Taking v – v = – ½ u gives: v = 5u/8, v = 9u/8, w = 9u/16
A B A B B
t1 = 8a/9u , t2 = 64a/171u, d = 4a/19
or taking v – v = – e u gives: v = (3 – e)u/4, v = 3(e + 1)u/4
A B A B
w = 3(e + 1)u/8, t1 = 4a/3(e + 1)u
B
t2 = 32ea/3(e + 1)(e + 9)u
d = 4ea/(e + 9) (max 8) | 5
4 | [9]
\includegraphics{figure_3}
Two perfectly elastic small smooth spheres $A$ and $B$ have masses $3m$ and $m$ respectively. They lie at rest on a smooth horizontal plane with $B$ at a distance $a$ from a smooth vertical barrier. The line of centres of the spheres is perpendicular to the barrier, and $B$ is between $A$ and the barrier (see diagram). Sphere $A$ is projected towards sphere $B$ with speed $u$ and, after the collision between the spheres, $B$ hits the barrier. The coefficient of restitution between $B$ and the barrier is $\frac{1}{4}$. Find the speeds of $A$ and $B$ immediately after they first collide, and the distance from the barrier of the point where they collide for the second time. [9]
\hfill \mbox{\textit{CAIE FP2 2010 Q3 [9]}}