2020 A/L Physics Past Paper · Question 49 (English Medium)

Answer: (4) $3\ \text{J}$

Given information

System: block $A$ ($2$ kg, $2\ \text{m s}^{-1}$), block $B$ ($6$ kg, at rest), two identical springs of negligible mass.

Frictionless horizontal surface.

What the question asks

We must find: the greatest elastic potential energy that can be stored in the two springs together during the collision.

Relevant theory

For a system of interacting objects, the momentum remains constant, provided no external force acts on the system.

If in a certain process the conservation of energy takes place only between kinetic energy and potential energy, then in that process the sum of the kinetic energy and the potential energy remains a constant. This is the principle of conservation of mechanical energy.

$$W = \frac{1}{2}mv^2$$

Required equations

$m_A u = (m_A + m_B)\,v$

$\tfrac{1}{2}m_A u^2 = \tfrac{1}{2}(m_A + m_B)v^2 + E_{\text{springs}}$

Applying it to this question

What stays constant: the total momentum of the system, and (kinetic energy + energy in the springs).

Initial state: before the springs touch.

Final state: greatest compression.

Reasoning and calculation

Why they are conserved: the surface is frictionless — no external horizontal force, so total momentum cannot change.

The springs are elastic and lossless — kinetic energy plus spring energy stays constant.

When is the compression greatest: while $A$ is faster than $B$, the gap between them keeps closing and the springs keep compressing.

The compression is greatest exactly when the two speeds become equal; at that instant the kinetic energy is least.

$4$

$m_A u = 2 \times 2$

$\tfrac{1}{2} \times 2 \times 2^2$

$0.5$

$v = \frac{4}{2 + 6}$

$1$

$\tfrac{1}{2} \times 8 \times 0.5^2$

$3$

$4 = 1 + E \Rightarrow E = 4 - 1$

Checks

Unit check: $\text{kg} \times (\text{m s}^{-1})^2 = \text{J}$ ✓.

$0.5 * mA * mB * u * u / (mA + mB)$

Second way: once the springs push the blocks apart again (elastic), $v_A = -1$, $v_B = +1\ \text{m s}^{-1}$ — momentum $2(-1) + 6(1) = 4$ ✓, K.E. $1 + 3 = 4$ J ✓.

In between, the greatest energy in the springs was $3$ J.

Conclusion

$E = 4 - 1 = 3\ \text{J}$

At greatest compression both blocks move at $0.5\ \text{m s}^{-1}$; K.E. drops from $4$ J to $1$ J; the springs hold $3$ J — the answer is (4).

Assuming all of $A$'s kinetic energy ($4$ J) goes into the springs.

Momentum is conserved, so the system cannot come to rest; even at greatest compression $\tfrac{1}{2}(m_A + m_B)v^2 = 1$ J of kinetic energy remains.