Answer :
a. To reach position A, the ball requires a minimum speed of approximately 8.86 m/s, while at position B, its speed matches the initial speed of 12.0 m/s.
b. The speed at A is 8.86 m/s, and at B, it remains 12.0 m/s.
To solve this problem, let's break it down into parts:
a)
i. To calculate the minimum speed v the ball must have in order to make it to position A, we can use the principle of conservation of mechanical energy. At position A, the gravitational potential energy is converted into kinetic energy.
The total mechanical energy ([tex]\( E_{\text{total}} \)[/tex]) at position A is the sum of the kinetic energy (KE) and the gravitational potential energy (PE):
[tex]\[ E_{\text{total}} = KE + PE \][/tex]
At position A, all the mechanical energy is in the form of kinetic energy. So, we have:
[tex]\[ E_{\text{total}} = KE = \frac{1}{2}mv^2 \][/tex]
At position A, the gravitational potential energy is given by:
PE = mgh
Where:
- m = mass of the ball
- v = velocity of the ball
- g = acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex])
- h = height (given as 4.0 m)
Setting the kinetic energy equal to the gravitational potential energy, we have:
[tex]\[ \frac{1}{2}mv^2 = mgh \][/tex]
Solving for v:
[tex]\[ v = \sqrt{2gh} \][/tex]
Now, we can substitute the given values to find the minimum speed v required for the ball to reach position A.
ii. To find the speed of the mass at position B, we can use the principle of conservation of mechanical energy again. At position B, all the mechanical energy is in the form of kinetic energy. So, the kinetic energy at B is the same as the total mechanical energy at A.
[tex]\[ KE_B = KE_A = \frac{1}{2}mv^2 \][/tex]
We can solve for the speed [tex]\( v_B \)[/tex] at position B using the given values of mass and height.
b) Given that v = 12.0 m/s, we'll calculate the speed at positions A and B using the same principles as above. We'll substitute v = 12.0 m/s into the equations for positions A and B to find the speeds. Let's do the calculations.
a)
i. To calculate the minimum speed v the ball must have in order to make it to position A, we'll use the equation:
[tex]\[ v = \sqrt{2gh} \][/tex]
Substituting the given values:
[tex]\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 4.0 \, \text{m}} \\\[ v = \sqrt{78.48} \\\[ v \approx 8.86 \, \text{m/s} \][/tex]
ii. To find the speed of the mass at position B, we can use the principle of conservation of mechanical energy. At position B, the total mechanical energy is the same as at position A. So, we have:
[tex]\[ KE_B = KE_A \\\[ \frac{1}{2}mv_B^2 = \frac{1}{2}mv^2 \][/tex]
Given v = 8.86 m/s, we have:
[tex]\[ \frac{1}{2}m \times v_B^2 = \frac{1}{2}m \times (8.86 \, \text{m/s})^2 \\\[ v_B^2 = (8.86 \, \text{m/s})^2 \\\[ v_B = 8.86 \, \text{m/s} \][/tex]
b)
Given v = 12.0 m/s, let's calculate the speed at positions A and B:
i. At position A:
[tex]\[ v_A = \sqrt{2gh} = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 4.0 \, \text{m}} \\\[ v_A \approx 8.86 \, \text{m/s} \][/tex]
ii. At position B:
[tex]\[ v_B = 12.0 \, \text{m/s} \][/tex]
So, the speed at position A is approximately 8.86 m/s and the speed at position B is 12.0 m/s.
