An apple with mass M is hanging at rest from the lower end of a light vertical rope. A dart of mass M/4 is shot vertically upwar

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3 months ago

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An apple with mass M is hanging at rest from the lower end of a light vertical rope. A dart of mass M/4 is shot vertically upwar

d, strikes the bottom of the apple, and remains embedded in it. If the speed of the dart is v₀ just before it strikes the apple, how high does the apple move upward because of its collision with the dart?

Physics

2 answers:

kicyunya [3.2K] · Answer 1 · 2026-03-24T22:21:18+03:00

Answer:

The apple rises to a height of $h=\frac{v_{0}^2}{50g}$ due to the dart's collision.

Explanation:

First we must determine the velocity acquired by the final mass after the collision occurs; for this, momentum conservation is utilized.

The movement is along a vertical axis, with only the dart initially in motion, and shortly after both masses being in motion together. We derive

$P_{i}=P_{f}\Leftrightarrow \frac{M}{4}v_{0}=\frac{5}{4}Mv$

From this, we can find the

velocity of the combined mass, , right after the collision (just after) $v$

$v=\frac{v_{0}}{5}$

Next, with this final velocity, we compute the maximum height that the mass could reach, using energy conservation. Following the collision, the combined mass possesses kinetic energy, and we consider this as our zero height (or zero potential energy), while at the peak, its kinetic energy becomes zero and potential energy changes. Thus, we formulate the following equality

$E_{i}=E_{f}\Leftrightarrow K_{i}=V_{f}\Leftrightarrow \frac{1}2}(\frac{5}{4}M)(\frac{v_{0}}{5})^2=(\frac{5}{4}M)gh$

From which we isolate for h to derive the answer to the posed question.

Which yields

$h=\frac{v_{0}^2}{50g}$

kicyunya [3.2K] · Answer 2 · 2026-03-24T22:19:20+03:00

Answer:

h = 1/50 v₀² g

Explanation:

This scenario illustrates conservation of momentum; the system comprises two bodies, hence forces during the collision are internal and momentum remains conserved.

Prior to the collision

po = 0 + (m/4) vo

Post collision

pf = (m + m/4) v

pf = (5m/4) v

m/4 vo = (5m/4) v

v = 1/5 vo

This indicates that after impact, the bodies travel at 1/5 of their initial velocity.

To determine the maximum height, we will apply the law of energy conservation.

At the peak post-collision

Em = K = ½ (5m / 4) v²

At the apex

Em = U = (5m/4) g h

½ (5m/4) v² = (5m/4) g h

h = ½ v² / g

h = ½ (1/5 v₀)² / g

h = 1/50 v₀² g