Because one side of the spoon is convex and the other side is concave, the spoon’s different sides reflect differently.
<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>
Answer:
Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]
Explanation:
We can solve this problem by using Newton's universal gravitation law.
In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m
![r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]](https://tex.z-dn.net/?f=r_%7Be%7D%20%3D%20distance%20earth%20to%20the%20astronaut%20%5Bm%5D.%5C%5Cr_%7Bm%7D%20%3D%20distance%20moon%20to%20the%20astronaut%20%5Bm%5D%5C%5Cr_%7Bt%7D%20%3D%20total%20distance%20%3D%203.84%2A10%5E8%5Bm%5D)
Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.
Mathematically this equals:
![F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]](https://tex.z-dn.net/?f=F_%7Bm%7D%20%3DG%2A%5Cfrac%7Bm_%7Bm%7D%2Am_%7Ba%7D%20%20%7D%7Br_%7Bm%7D%20%5E%7B2%7D%20%7D%20%5C%5Cwhere%3A%5C%5CG%20%3D%20gravity%20constant%20%3D%206.67%2A10%5E%7B-11%7D%5B%5Cfrac%7BN%2Am%5E%7B2%7D%20%7D%7Bkg%5E%7B2%7D%20%7D%20%5D%20%5C%5Cm_%7Be%7D%3D%20earth%27s%20mass%20%3D%205.98%2A10%5E%7B24%7D%5Bkg%5D%5C%5C%20m_%7Ba%7D%3D%20astronaut%20mass%20%3D%20100%5Bkg%5D%5C%5Cm_%7Bm%7D%3D%20moon%27s%20mass%20%3D%207.36%2A10%5E%7B22%7D%5Bkg%5D)
When we match these equations the masses cancel out as the universal gravitational constant
To solve this equation we have to replace the first equation of related with the distances.
Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.
![r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]](https://tex.z-dn.net/?f=r_%7Bm1%2C2%7D%3D%5Cfrac%7B-b%2B-%20%5Csqrt%7Bb%5E%7B2%7D-4%2Aa%2Ac%20%7D%20%20%7D%7B2%2Aa%7D%5C%5C%20%20where%3A%5C%5Ca%3D80.25%5C%5Cb%3D768%2A10%5E%7B6%7D%20%5C%5Cc%20%3D%20-1.47%2A10%5E%7B17%7D%20%5C%5Creplacing%3A%5C%5Cr_%7Bm1%2C2%7D%3D%5Cfrac%7B-768%2A10%5E%7B6%7D%2B-%20%5Csqrt%7B%28768%2A10%5E%7B6%7D%29%5E%7B2%7D-4%2A80.25%2A%28-1.47%2A10%5E%7B17%7D%29%20%7D%20%20%7D%7B2%2A80.25%7D%5C%5C%5C%5Cr_%7Bm1%7D%3D%2038280860.6%5Bm%5D%20%5C%5Cr_%7Bm2%7D%3D-2.97%2A10%5E%7B17%7D%20%5Bm%5D)
We work with positive value
rm = 38280860.6[m] = 38280.86[km]
<u>Second part</u>
<u />
The distance between the Earth and this point is calculated as follows:
re = 3.84 108 - 38280860.6 = 345719139.4[m]
Now the acceleration can be found as follows:
![a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]](https://tex.z-dn.net/?f=a%20%3D%20G%2A%5Cfrac%7Bm_%7Be%7D%20%7D%7Br_%7Be%7D%20%5E%7B2%7D%20%7D%20%5C%5Ca%20%3D%206.67%2A10%5E%7B11%7D%20%2A%5Cfrac%7B5.98%2A10%5E%7B24%7D%20%7D%7B%28345.72%2A10%5E%7B6%7D%29%5E%7B2%7D%20%20%7D%20%5C%5Ca%3D3.33%2A10%5E%7B19%7D%20%5Bm%2Fs%5E2%5D)
Answer:
Explanation:
(a) When the plate starts to spin:
Its angular velocity increases, so the angular acceleration is non zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
(b) When the plate rotates at constant angular velocity:
Its angular velocity is constant so the angular acceleration is zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
(c) When the plate sows to halt:
Its angular velocity decreases, so the angular acceleration is non zero( but negative). As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.
