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Law Incorporation [45]
3 years ago
13

A banana is fired from a cannon on the ground just as a monkey starts to fall from a tree. The monkey catches the banana in mida

ir as it falls to the ground. What was definitely true about the banana and monkey just before the monkey caught the banana?
A. The banana and monkey traveled the same distance from their initial positions.
B. The banana and monkey had the same vertical velocity.
C. The banana and monkey had the same acceleration.
D. All of the above.
Physics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Here banana is fired from the canon towards the monkey with certain velocity

while at the same instant when banana is fired money starts to fall from the tree

So here when monkey is in the mid air he catches the banana

It is possible because in the same interval of time when the monkey cover some distance from the tree towards ground then in the same time banana cover the distance from ground towards monkey

So they meet at some point in the mid air

So here since both are moving free in air so they both must have same vertical acceleration

this is acceleration due to gravity

so here correct answer will be

C. The banana and monkey had the same acceleration.

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A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.
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Approximately 1.6\times 10^{3}\; \rm N.

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By the Impulse-Momentum Theorem, the change in this woman's momentum  will be equal to the impulse that is applied to her.

The momentum p of an object is equal to the product of its mass m and velocity v. That is: p = m \cdot v.

Let v(\text{before}) and v(\text{after}) represent the velocity of the woman before and after the landing. Let m represent the woman's mass.

  • The woman's momentum before the landing would be m \cdot v(\text{before}).
  • The woman's momentum after the landing would be m \cdot v(\text{after}).

Therefore, the change in this woman's momentum would be:

\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let F represent the average force on the woman. The impulse on her during the landing would be F \cdot t.

Apply the Impulse-Momentum Theorem.

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Impulse is equal to the change in momentum:

F \cdot t = m \cdot (v(\text{after})- v(\text{before})).

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, v(\text{after}) = 0\; \rm m \cdot s^{-1}.

\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}.

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