Ball Motion Analysis

This question asks us to identify the correct relationship between kinetic energy (KE) and potential energy (PE) for a ball of mass M undergoing vertical motion. The ball is thrown upwards from point A, reaches a highest point B, and then returns to A. We need to analyze the energy at points A and B.

Energy Principles Explained

To solve this, we use fundamental physics concepts:

• Kinetic Energy (KE): This is the energy associated with the motion of an object. It depends on the object's mass and velocity. The formula is $KE = \frac{1}{2}mv^2$.
• Potential Energy (PE): This is the energy stored in an object due to its position. In this case, we consider gravitational potential energy, relative to a reference point. The formula is $PE = mgh$, where $h$ is the height above the reference point.
• Conservation of Mechanical Energy: Assuming no energy loss due to air resistance or other non-conservative forces, the total mechanical energy (the sum of KE and PE) of the ball remains constant throughout its flight. Mathematically, $E_{total} = KE + PE = \text{constant}$.

Energy at Point A (Start)

Point A is the starting position from where the ball is thrown upwards.

• We usually set the potential energy at the ground level or the starting point to zero. So, let $PE_A = 0$.
• The ball is thrown upwards with some initial velocity, which means it has kinetic energy at point A. Let this be $KE_A$. Since it's thrown upwards, $KE_A > 0$.
• The total mechanical energy at point A is $E_A = KE_A + PE_A = KE_A + 0 = KE_A$.

Energy at Point B (Highest Point)

Point B is the highest point the ball reaches.

• At the very peak of its trajectory (point B), the ball momentarily stops before it starts falling back down. This means its vertical velocity is zero ($v = 0$).
• Consequently, the kinetic energy at point B is zero: $KE_B = \frac{1}{2}M(0)^2 = 0$.
• Since point B is at a certain height ($h$) above point A, the ball possesses gravitational potential energy at B. Let this be $PE_B = Mgh$. Since $h > 0$, $PE_B > 0$.
• The total mechanical energy at point B is $E_B = KE_B + PE_B = 0 + PE_B = PE_B$.

Applying Energy Conservation

According to the principle of conservation of mechanical energy:

$E_A = E_B$

Substituting the expressions for energy at points A and B:

$KE_A + PE_A = KE_B + PE_B$

Now, plug in the values we established ($PE_A = 0$ and $KE_B = 0$):

$KE_A + 0 = 0 + PE_B$

This simplifies the equation to:

$KE_A = PE_B$

Evaluating the Options

Let's compare our derived relationship ($KE_A = PE_B$) with the given options:

• Option 1: Kinetic Energy at A = Potential Energy at B - This matches our derived equation $KE_A = PE_B$. This is the correct relationship.
• Option 2: Kinetic Energy at A = Potential Energy at A - This implies $KE_A = 0$, because we defined $PE_A = 0$. This is incorrect, as the ball needs initial velocity to go up.
• Option 3: Kinetic Energy at B = Potential Energy at B - This implies $0 = PE_B$ (since $KE_B=0$). This is incorrect because point B is at a height above A, thus $PE_B$ cannot be zero.
• Option 4: Kinetic Energy at B = Kinetic Energy at A - This implies $0 = KE_A$ (since $KE_B=0$). This is incorrect, as the ball is thrown with an initial velocity.

Thus, the kinetic energy the ball possesses at the moment it is thrown from point A is converted entirely into potential energy when it reaches its maximum height at point B.