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irina1246 [14]
3 years ago
15

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

en the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.
2. If an object of mass m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor of ___. The new velocity will be ___v.

3. If an object of mass 3m collides and velocity v collides inelastically with an object of mass 4m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.

4. If an object of mass 5m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, then the amount of total system mass in motion will increase by a factor of ___ and the velocity of the system will decrease by a factor if ___. The new velocity will be ___v.
Physics
1 answer:
Archy [21]3 years ago
3 0

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f  + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

  final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4  ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a)  final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5  ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a)  final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀  ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a)  final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

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Answer:

Speed:

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Wavelength:

4.24x10^{-7}m

Frequency:

4.74x10^{14}Hz

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

n = \frac{c}{v} (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

v = \frac{c}{n}

v = \frac{3x10^{8}m/s}{1.490}

v = 2.01x10^{8}m/s

Hence, the speed of the laser has a value of 2.01x10^{8}m/s

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.  

To determine the frequency it can be used the following equation

c = \nu \cdot \lambda  (2)

Where c is the speed of light, \nu is the frequency and \lambda is the wavelength

Then, \nu wil be isolated from equation 2.

\nu = \frac{c}{\lambda}  (3)

Before using equation 3 it is necessary to express \lamba in units of meters.

\lambda = 632.8nm . \frac{1m}{1x10^{9}nm} ⇒ 6.328x10^{-7}m

\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}

\nu = 4.74x10^{14}s^{-1}

\nu = 4.74x10^{14}Hz

Hence, the frequency of the laser has a value of 4.74x10^{14}Hz

Wavelength:

To determine the wavelength it can be used:

v = \nu \cdot \lambda

\lambda = \frac{v}{\nu}

Where v is the speed of the laser through the polystyrene.

\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}

\lambda = 4.24x10^{-7}m

Hence, the wavelength of the laser has a value of 4.24x10^{-7}m

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