Challenging +1.2 This is a standard two-stage mechanics problem combining conservation of momentum (for the collision) and energy conservation (for the spring compression). While it requires careful algebraic manipulation and involves a parameter k, the approach is straightforward: find velocity after collision using momentum, then equate kinetic energy to elastic potential energy at maximum compression. The setup is clear and the techniques are standard A-level Further Maths mechanics content.
A light spring \(AB\) has natural length \(a\) and modulus of elasticity \(5mg\). The end \(A\) of the spring is attached to a fixed point on a smooth horizontal surface. A particle \(P\) of mass \(m\) is attached to the end \(B\) of the spring. The spring and particle \(P\) are at rest on the surface.
Another particle \(Q\) of mass \(km\) is moving with speed \(\sqrt{4ga}\) along the horizontal surface towards \(P\) in the direction \(BA\). The particles \(P\) and \(Q\) collide directly and coalesce. In the subsequent motion the greatest amount by which the spring is compressed is \(\frac{2}{3}a\).
Find the value of \(k\). [6]
Loss in KE = Gain in EPE: 1 m ( k+1 ) v2 = 1 × 5mg × a
Answer
Marks
Guidance
2 2 a 5
M1
Energy equation, LHS correct, EPE dimensionally correct.
Substitute for v and rearrange to form quadratic equation in k
Answer
Marks
20k2 =1+k
M1
1
k =
Answer
Marks
4
A1
6
Answer
Marks
Guidance
Question
Answer
Marks
Question 2:
2 | At the collision of P and Q: ( m+km ) v=kmu | M1 | Momentum conserved, allow missing k on RHS.
k 4ga
So v=
( 1+k ) | A1
2
1 5mg a mga
EPE = × × =
2 a 5 10 | B1
2
Loss in KE = Gain in EPE: 1 m ( k+1 ) v2 = 1 × 5mg × a
2 2 a 5 | M1 | Energy equation, LHS correct, EPE dimensionally correct.
Substitute for v and rearrange to form quadratic equation in k
20k2 =1+k | M1
1
k =
4 | A1
6
Question | Answer | Marks | Guidance
A light spring $AB$ has natural length $a$ and modulus of elasticity $5mg$. The end $A$ of the spring is attached to a fixed point on a smooth horizontal surface. A particle $P$ of mass $m$ is attached to the end $B$ of the spring. The spring and particle $P$ are at rest on the surface.
Another particle $Q$ of mass $km$ is moving with speed $\sqrt{4ga}$ along the horizontal surface towards $P$ in the direction $BA$. The particles $P$ and $Q$ collide directly and coalesce. In the subsequent motion the greatest amount by which the spring is compressed is $\frac{2}{3}a$.
Find the value of $k$. [6]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q2 [6]}}