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Nostrana [21]
3 years ago
5

The football players collide head-on in midair while trying to catch a thrown football. The first player is 95.0 kg and has an i

nitial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of –3.50 m/s. What is their velocity just after impact if they cling together?
Physics
1 answer:
sukhopar [10]3 years ago
3 0

Answer:

v=0.8m/s

Explanation:

To solve the problem, we can use collisions theory from classical physics

If we analyze the players before and after the collision we got:

p_{1}=p_{2}

Since the cling together

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

From the exercise we know that the first player is 95.0 kg and has an initial velocity of 6.00 m/s, while the second player is 115 kg and has an initial velocity of –3.50 m/s.

v=\frac{m_{1}v_{1}+m_{2}v_{2}}{(m_{1}+m_{2})}=\frac{(95kg)(6m/s)+(115kg)(-3.5m/s)}{(95+115)kg}

v=0.797m/s=0.8m/s

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A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t
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Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

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