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True [87]
3 years ago
9

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

30 degrees from the horizontal and the 3kg object is deflected 315 degrees from the horizontal. Determine the velocity of each ball after the collision.
Physics
1 answer:
gavmur [86]3 years ago
3 0

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected  = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta

Put the value into the fomrula

5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45

15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}

15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}....(I)

Along y -axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta

Put the value into the formula

0+0=5\times v_{1}\sin30-3\times v_{2}\sin45

\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0...(II)

From equation (I) and (II)

v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}

v_{1}=2.19\ m/s

Put the value of v₁ in equation (I)

\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0

v_{2}=\dfrac{5.475\times\sqrt{2}}{3}

v_{2}=2.58\ m/s

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

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Answer:

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Explanation:

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         v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

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where n is the index of refraction of the material medium.

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           f =\frac{c }{\lambda_o}  

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let's calculate

           v = \frac{450 \ 10^{-9} }{1.00029}  \ 6.667 \ 10^{14}450 10-9 / 1,00029 6,667 1014

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The calculated mutual inductance is 8.544 x 10⁻⁵ H.

Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.

Length of the solenoid= 5.0 cm

Area of cross-section=1.0 cm²

no of spaced turns=300 turns

turns of insulated wire=180 turns

Mutual inductance (M) = μ₀μr N1N2 A/ L

                                     =(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5

                                     =79.12 x 10⁻¹⁰ x 54000 / 5

                                     =8.544 x 10⁻⁵  H

hence, the mutual inductance is 8.544 x 10⁻⁵ H.

Learn more about Mutual inductance here-

brainly.com/question/14014588

#SPJ4

                                   

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