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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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Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
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