All categories

3 years ago

Elastic collision formula find each velocity

1 answer:

7 0

We know the initial velocity of the golf ball and its mass, but we don't know the final velocitiesof either ball, and the trick to make these calculations go faster for an elastic collision is to use this equation, which says the initial velocity of one of the objects before the collision

Hoped this helped!

You might be interested in

A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

E = Vo / D

E = 600 / 0.3

E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

Fe = E*q

Fe = (2,000 V / m) * ( 3*10^-5 C)

Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

Fe = m*a

a = Fe / m

a = 0.06 / 0.0004

a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

s = C - A

s = ( 0.25 , 12 ) - ( 0.05 , 12 )

s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

vf^2 = vi^2 + 2*a*s

vf^2 = 0 + 2*0.2*150

vf = √60

vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

vf - vi = a*t

t = ( 7.746 - 0 ) / 150

t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

$Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]$

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

$E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\$

- Again, apply the Newton's second law of motion and determine the acceleration (a):

Fe = m*a

a = Fe / m

$a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}$

- Where the acceleration is rate of change of velocity "dv/dt":

$\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\$

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

$c = \frac{A*eo}{d}$

Where, eo = 8.854 * 10^-12 .... permittivity of free space.

$K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\$

- The differential equation turns out ot be:

$\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\$

- Separate the variables the integrate over the interval :

t : ( 0 , t )

v : ( 0 , vf )

Therefore,

$\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )$

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

$vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s$

- The final velocity of particle while charging the capacitor would be:

vf = 7.264 m/s ... slightly less for the fully charged capacitor

7 0

3 years ago

How much work is accomplished when a force of 300N pushes a box across the floor for a distance of 100 meters?

Nesterboy [21]

So the correct ans is B.

hope it helps u.

5 0

4 years ago

Read 2 more answers

Two identical automobiles are

Phantasy [73]

Answer:

i think that it is b but I could be wrong

3 0

3 years ago

A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

3 years ago

Read 2 more answers

You are an employee of the city of Chicago. A city official declares that all cables holding traffic lights in the city should b

Irina18 [472]

If you have ever lived in Chicago, then the answer might very well be C.
Newton's Second law states F = ma. This problem has nothing to do with acceleration.
Newton's Third law is the familiar action/reaction law. If the poles are anchored well enough and have a flexural strength greater than the tension exerted on them by the wire; then D is just.
That leaves B. There's no problem all we have to do is to increase the horizontal tension in the cable.

5 0

3 years ago