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Fantom [35]
3 years ago
10

Suppose two masses labelled m1 and m2, are speeding toward each other at time t=0s. The first mass 1 has a mass and speed has va

lues of m1=3 kg and v1,i= 2 m/s (to the right). Mass 2 has a mass and speed of m2 = 5kg with an initial speed of v2,I = 5 m/s (to the left). Then at some time t there is an elastic collision between the two masses. What is the final speed and direction of m1 and m2 after the collision.
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

The velocity of the first mass is 19.9 m/s to the left

The velocity of the second mass is 8.14 m/s to the left

Explanation:

In an elastic collision, both momentum and kinetic energy is conserved.

\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_{11}^2 + \frac{1}{2}m_2v_{22}^2\\m_1v_1 + m_2v_2 = m_1 v_{11} + m_2 v_{22}

There is a lot of elaboration to solve these two equations, but substituting the values given in the question will ease the hard work.

\frac{1}{2}(3)(2)^2 + \frac{1}{2}(5)(5)^2 = \frac{1}{2}(3)v_{11}^2 + \frac{1}{2}(5)v_{22}^2\\(3)(2) - (5)(5) = (3) v_{11} + (5) v_{22}\\6 + \frac{125}{2} = \frac{1}{2}(3)v_{11}^2 + \frac{1}{2}(5)v_{22}^2\\-19 = 3 v_{11} + 5 v_{22}\\v_{11}^2 = (\frac{-19 - 5v_{22}}{3})^2\\{\rm Plugging ~this ~into~the~kinetic~energy~equation~gives:}\\\frac{137}{2} = \frac{3}{2}(\frac{-19-5v_{22}}{3})^2 + \frac{5}{2}v_{22}\\137 = \frac{361 + 190v_{22} + 25v_{22}^2}{3} + \frac{5}{2}v_{22}

Rearranging the equations gives

137 = \frac{722 + 380v_{22} + 50v_{22}^2 + 15v_{22}}{6}\\822 = 722 + 395v_{22} + 50v_{22}^2\\50v_{22}^2 + 395v_{22} - 100 = 0\\10v_{22}^2 + 79v_{22} - 20 = 0

Solving this equation quadratically gives the velocity of the second mass:

v_{22} = -8.14~{\rm or}~0.24

There are two roots to the quadratic equation, but we intuitively know that the bigger mass with the higher initial velocity will have the same direction after the collision.

Therefore, the final speed of the second mass is 8.14 m/s to the left.

Now, it is easy to calculate the velocity of the first mass.

-19 = (3)v_{11} -(-8.14)5\\ v_{11} = -19.9

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dalvyx [7]
The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
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4 0
3 years ago
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ruslelena [56]

Answer:

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3 0
3 years ago
Read 2 more answers
Help show how you got answers
lapo4ka [179]

The speed at which a wave of frequency of 14 Hz and a wavelength of 3 m will travel is  42 m / s

v = λ / T

f = 1 / T

v = f λ

v = Wave speed

λ = Wavelength

T = Time period

f = Frequency

f = 14 Hz

λ = 3 m

v = 14 * 3

v = 42 m / s

Frequency is the number of time an event occurs repeatedly in an unit amount of time. Its unit is hertz which in terms of SI unit is s^{-1}.

Therefore, the speed at which the wave will travel is 42 m / s

To know more about frequency

brainly.com/question/16861358

#SPJ1

4 0
1 year ago
Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart
Dahasolnce [82]

Answer:F=\frac{(M+m)g}{\mu _s}

Explanation:

Given

Trolley of mass M is free to roll without friction

coefficient of friction between trolley and mass m is \mu _s

Force F is applied on mass m

Acceleration of the system is

a=\frac{F}{M+m}

friction Force will balance weight of block

friction force=\mu _sN

N=ma

\mu _sN=mg

\mu _sma=mg

\mu _s=\frac{g}{a}

F=\frac{(M+m)g}{\mu _s}

6 0
3 years ago
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