Kinematics

1. Analyze the given information:

• Initial position: Origin (0, 0)
• Initial velocity: ms^-1
• Acceleration: ms^-2
• x-coordinate at a certain instant: 16 m

2. Find the time:

We can use the equation of motion for the x-direction:

Since the initial velocity is in the y-direction, . The x-component of acceleration is m/s². We are given m. Substituting these values:

s (We take the positive value of time)

3. Find the y-coordinate:

Now we can use the equation of motion for the y-direction:

The initial velocity in the y-direction is m/s, the y-component of acceleration is m/s², and we found s. Substituting these values:

m

The correct answer is (D) 24 m.

1. Understand the Graph:
Maximum acceleration of at ,
Decreasing linearly to at .
This suggests that the acceleration decreases uniformly from to over the time interval to .

2. Use the Formula for Maximum Velocity:
The velocity is the integral of acceleration with respect to time: $ t = 10 \, \text{s} a = 8 \, \text{m/s}^2 40 \, \text{m/s}$

1. Understanding the Graph:

The first part of the graph has a positive slope, indicating increasing velocity. This corresponds to a body undergoing uniform acceleration (since the velocity is changing at a constant rate).

The second part of the graph is horizontal, indicating that the velocity is constant. This corresponds to uniform motion (no acceleration).

2. Key Concepts:

1. Uniform Acceleration: - Occurs when the velocity changes at a constant rate. - In a v-t graph, uniform acceleration is represented by a straight line with a non-zero slope (positive or negative). - The area under this line gives the displacement during that time interval.

2. Uniform Motion: - Occurs when the velocity is constant. - In a v-t graph, uniform motion is represented by a horizontal line (slope = 0). - The area under the line gives the displacement over time, but since the velocity is constant, the displacement is directly proportional to time.

3. Displacement During Uniform Acceleration and Uniform Motion:

- Displacement during Uniform Acceleration (first part of the graph): - The displacement during uniform acceleration can be calculated by finding the area under the curve. - The area under a straight line with a positive slope (velocity vs. time) gives the displacement. - The formula for displacement when the acceleration is uniform is: $ $

4. Final Answer

Therefore, the displacement during uniform acceleration is greater than that during uniform motion.

Step 1: Kinetic energy of proton (Non-relativistic)

Step 2: Solve for

Step 3: Momentum of photon ( )

Energy of photon

Momentum of photon

Step 4: Calculate

Step 5: Match to Option

Option (D) is .

Final Answer: The final answer is

Given:

• Mass of the ball, kg
• Initial height, m
• Energy lost after collision = 50%
• Acceleration due to gravity, m/s²

Step 1: Use Energy Conservation

Let initial velocity be . The total mechanical energy before hitting the ground:

After hitting the ground, the ball loses 50% of its energy and rises back to the same height, meaning its remaining energy is:

Step 2: Express in Terms of Initial Velocity

Since the ball reaches the same height after losing 50% energy, the energy left is equal to its potential energy at height :

Solving for :

Answer: The initial velocity of the ball is 14 m/s (Option A).

Given:

• Initial velocity:
• Final velocity after seconds:
• Acceleration: (constant)

Step 1: Using Kinematic Equation

We use the third equation of motion:

Solving for (displacement):

Step 2: Average Velocity Formula

Average velocity is given by:

Step 3: Conclusion

The correct answer is:

Answer: The correct option is C.

By applying the principle of conservation of mechanical energy, the potential energy of the body is completely converted into kinetic energy during the fall.
Using the equation we solve for , which gives .