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Kay
1 day ago
5

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 [2.2K]1 day ago
8 0

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 [2.2K]1 day ago
3 0

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

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Answer:

Electric flux is calculated as \phi=31562.63\ Nm^2/C

Explanation:

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The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

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A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
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Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

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Now, the potential for the smaller droplets is given as:

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