Hlobsile Themba
Oct 2
Calculate the maximum kinetic energy that marble B gains during its motion.
Calculation
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ChatGPT
Oct 2
To calculate the maximum kinetic energy that marble B gains during its motion, we need to first determine its maximum speed.
Given that marble B starts from rest and rolls down a ramp with a height of 2 meters, we can use the principle of conservation of energy to find its maximum speed. The potential energy at the top of the ramp is converted into kinetic energy at the bottom of the ramp.
The potential energy at the top of the ramp is given by:
PE = mgh
where m is the mass of the marble, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (2 meters).
PE = 0.1 kg * 9.8 m/s^2 * 2 m
PE = 1.96 J
This potential energy is converted into kinetic energy at the bottom of the ramp, given by:
KE = 0.5 * m * v^2
where v is the speed of the marble at the bottom of the ramp.
Setting the potential energy equal to the kinetic energy, we have:
1.96 J = 0.5 * 0.1 kg * v^2
v^2 = 1.96 J / (0.5 * 0.1 kg)
v^2 = 39.2 m^2/s^2
v = √39.2 m/s
v ≈ 6.26 m/s
Now that we have found the maximum speed of marble B, we can calculate its maximum kinetic energy:
KE = 0.5 * 0.1 kg * (6.26 m/s)^2
KE = 0.5 * 0.1 kg * 39.2 m^2/s^2
KE = 1.96 J
Therefore, the maximum kinetic energy that marble B gains during its motion is 1.96 Joules.
Given that marble B starts from rest and rolls down a ramp with a height of 2 meters, we can use the principle of conservation of energy to find its maximum speed. The potential energy at the top of the ramp is converted into kinetic energy at the bottom of the ramp.
The potential energy at the top of the ramp is given by:
PE = mgh
where m is the mass of the marble, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (2 meters).
PE = 0.1 kg * 9.8 m/s^2 * 2 m
PE = 1.96 J
This potential energy is converted into kinetic energy at the bottom of the ramp, given by:
KE = 0.5 * m * v^2
where v is the speed of the marble at the bottom of the ramp.
Setting the potential energy equal to the kinetic energy, we have:
1.96 J = 0.5 * 0.1 kg * v^2
v^2 = 1.96 J / (0.5 * 0.1 kg)
v^2 = 39.2 m^2/s^2
v = √39.2 m/s
v ≈ 6.26 m/s
Now that we have found the maximum speed of marble B, we can calculate its maximum kinetic energy:
KE = 0.5 * 0.1 kg * (6.26 m/s)^2
KE = 0.5 * 0.1 kg * 39.2 m^2/s^2
KE = 1.96 J
Therefore, the maximum kinetic energy that marble B gains during its motion is 1.96 Joules.
0
Google Gemini AI
Oct 2
Given:
- Mass of marble A: m
- Mass of marble B: 2m
- Velocity of marble A: v
- Velocity of marble B after collision: 0
Solution:
1. Conservation of Momentum:
Since the system is isolated, the total momentum before and after the collision remains the same:
```
mv + 2m(0) = mv_A + 2mv_B
```
Solving for the velocity of marble A after the collision (v_A):
```
v_A = (2/3)v
```
2. Conservation of Kinetic Energy:
The total kinetic energy before the collision (K_i) is:
```
K_i = (1/2)mv^2 + (1/2)(2m)(0)^2 = (1/2)mv^2
```
The total kinetic energy after the collision (K_f) is:
```
K_f = (1/2)m(v_A)^2 + (1/2)(2m)(v_B)^2
```
Since marble B has zero velocity after the collision, its kinetic energy is zero:
```
K_f = (1/2)m(v_A)^2
```
3. Maximum Kinetic Energy of Marble B:
The maximum kinetic energy that marble B can gain is when all the kinetic energy of marble A is transferred to it. This occurs when marble A stops completely after the collision (v_A = 0).
Substituting v_A = 0 into the equation for K_f, we get:
```
K_f = (1/2)m(0)^2 = 0
```
Therefore, the maximum kinetic energy that marble B gains is zero.
- Mass of marble A: m
- Mass of marble B: 2m
- Velocity of marble A: v
- Velocity of marble B after collision: 0
Solution:
1. Conservation of Momentum:
Since the system is isolated, the total momentum before and after the collision remains the same:
```
mv + 2m(0) = mv_A + 2mv_B
```
Solving for the velocity of marble A after the collision (v_A):
```
v_A = (2/3)v
```
2. Conservation of Kinetic Energy:
The total kinetic energy before the collision (K_i) is:
```
K_i = (1/2)mv^2 + (1/2)(2m)(0)^2 = (1/2)mv^2
```
The total kinetic energy after the collision (K_f) is:
```
K_f = (1/2)m(v_A)^2 + (1/2)(2m)(v_B)^2
```
Since marble B has zero velocity after the collision, its kinetic energy is zero:
```
K_f = (1/2)m(v_A)^2
```
3. Maximum Kinetic Energy of Marble B:
The maximum kinetic energy that marble B can gain is when all the kinetic energy of marble A is transferred to it. This occurs when marble A stops completely after the collision (v_A = 0).
Substituting v_A = 0 into the equation for K_f, we get:
```
K_f = (1/2)m(0)^2 = 0
```
Therefore, the maximum kinetic energy that marble B gains is zero.
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