2020 A/L Physics Past Paper · Question 23 (English Medium)

Answer: (4) Diagram (4)

What the question asks

We must find: the diagram showing the path of the half that does not fall straight down, after the object explodes at the highest point P.

Relevant theory

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

However, within the system, the interactions between the objects cause transfer of momentum between them. But the total momentum stays constant.

$m u = \tfrac{m}{2}\cdot 0 + \tfrac{m}{2}u'$

$0 = \tfrac{m}{2}(-v) + \tfrac{m}{2}v'$

$u' = 2u,\quad v' = +v$

Applying it to this question

Shown in the diagram: an object of mass $m$ moving along a projectile path.

Its highest point P; there it explodes into two equal halves ($\tfrac{m}{2}$ each).

One half follows the downward arrow (initial speed $v$).

The path without the explosion is the dashed line.

Interactions: the forces of the explosion are internal forces between the two halves.

Over the very short explosion time, the impulse of gravity is negligible.

After the explosion, only gravity acts on each half.

Reasoning and calculation

Read from the diagram: P is the highest point — so the vertical speed there is zero, the velocity is purely horizontal at $u$ (to the right).

Half 1's velocity is vertical at $v$ downward; its horizontal speed is zero.

Rule that applies: the principle of conservation of momentum.

The total momentum just before and just after the explosion, both horizontal and vertical, is equal.

Applying it — horizontal:

$mu = \tfrac{m}{2}u'$ ⇒ $u' = 2u$

vertical:

$0 = -\tfrac{m}{2}v + \tfrac{m}{2}v'$ ⇒ $v' = +v$ (upward)

So half 2 first rises above P.

Its horizontal speed is doubled, and its flight time is longer than the remaining flight time of the un-exploded path — so it lands far beyond the dashed path.

Reading the options

Expected path: rising upward from P, moving forward at double the speed.

Landing far beyond where the dashed line would reach the ground — diagram (4).

Why the other options fail

The other half also moves straight down — the horizontal momentum $mu$ disappears.

It moves backward — the direction of horizontal momentum is reversed.

It does not rise above P — an upward momentum is needed to balance the downward half's vertical momentum.

It rises, but the landing distance is too short — the horizontal speed $2u$ and the long flight time give a much greater distance.

Checks

Physical plausibility: as long as both halves are in the air, their centre of mass keeps moving along the dashed path.

The external force (gravity) has not changed.

With numbers (a model: $u = 6$, launch vertical speed $14$, $v = 6\ \text{m s}^{-1}$):

range without the explosion is $16.8$ m.

Half 2 lands at about $33.9$ m ✓.

Conclusion

$u' = 2u$ forward, $v' = v$ upward: the other half first rises, then lands far beyond the dashed path — the answer is (4).

Thinking only horizontal momentum is conserved in the explosion.

Momentum is conserved in every direction: the vertical momentum at P is zero, so if one half goes down, the other goes up.