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kolezko [41]
3 years ago
12

Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and

the other's was 3.6 m/s in the opposite direction, what will be their speeds and directions after the collision
Physics
1 answer:
My name is Ann [436]3 years ago
6 0
<span>The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
 The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s

   There are two principles involved here
 Conservation of momentum and conservation of energy.
  I'll use the following variables
 a0, a1 = velocity of ball a (before and after collision)
  b0, b1 = velocity of ball b (before and after collision)
 m = mass of each ball.

   For conservation of momentum, we can create this equation:
 m*a0 + m*b0 = m*a1 + m*b1
 divide both sides by m and we get:  
a0 + b0 = a1 + b1

   For conservation of energy, we can create this equation:
 0.5m(a0)^2 + 0.5m(b0)^2 = 0.5m(a1)^2 + 0.5m(b1)^2
 Once again, divide both sides by 0.5m to simplify
 a0^2 + b0^2 = a1^2 + b1^2

   Now let's get rid of a0 and b0 by assigned their initial values. a0 will be 2, and b0 will be -3.6 since it's moving in the opposite direction.
 a0 + b0 = a1 + b1
 2 - 3.6 = a1 + b1
 -1.6 = a1 + b1
 a1 + b1 = -1.6

   a0^2 + b0^2 = a1^2 + b1^2
 2^2 + -3.6^2 = a1^2 + b1^2
 4 + 12.96 = a1^2 + b1^2
 16.96 = a1^2 + b1^2
 a1^2 + b1^2 = 16.96

   The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.</span>
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The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co
Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density = 8.4\times10^{28}\ m^3

Current i = 4.6\times10^{18}\ electron/s

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

v_{d}=\dfrac{I}{neA}

v_{d}=\dfrac{Ne}{tne\times\pi r^2}

Put the value into the formula

v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}

v_{d}=4.842\times10^{-5}\ m/s

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Using formula for time

v_{d}=\dfrac{l}{t}

t=\dfrac{l}{v_{d}}

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3 years ago
Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it
makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

 Here, we have been given the height through which the ball drops;

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The right motion equation is;

      V²  = U² + 2gh

V is the final velocity

U is the initial velocity  = 0

g is the acceleration due to gravity  = 9.8m/s²

h is the height

  Now insert the parameters and solve;

       V² = 0² + 2 x 9.8 x 12.6

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Check the correctness of formula t=2π √m/k, dimensionally​
pantera1 [17]

Hi there! Lets see!

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Therefore:

\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}}  =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s

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