Problem 20 - Applied Mathematics

Question

A particle of mass \(m\) is thrown vertically upwards with speed \(u\) from a point A on the ground. Simultaneously an identical particle is thrown vertically downwards also with speed \(u\) from a point B vertically above \(\mathrm{A}\) and at a height \(h\) above the ground \(\left(h<4 u^{2} / g\right)\). On impact the particles adhere and move subsequently as a single particle. Calculate the loss in kinetic energy caused by the impact and the speed of the combined particle on reaching the ground.

Step-by-Step Solution

Answer

The loss in kinetic energy is \(mu^2\). Speed of the combined particle on reaching the ground is \(\sqrt{2gh}\)

Step 1 - Calculation of Time of Flight for Upward Particle

Calculate the time taken by the particle thrown upwards with speed \(u\) to reach the point B. Use the equation: \[ t = \frac{\u}{g} \text{where g is the acceleration due to gravity} \]

Step 2 - Calculation of Distance Covered by Downward Particle

Calculate the distance covered by the particle thrown downwards from point B by the same time it reaches point A. Use the equation of motion: \[ h = ut+\frac{1}{2}gt^{2} \]

Step 3 - Conservation of Momentum on Impact

Apply conservation of momentum for the collision: \[ mu-mu = ( m + m )v \] Here, the velocity of the system immediately after impact is zero \(v=0\).

Step 4 - Calculation of Kinetic Energy Before Collision

Calculate the initial kinetic energy of the system before the collision. Use the formula for kinetic energy: \[ KE_{initial} = \frac{1}{2} m u^2 + \frac{1}{2} m u^2 = mu^2 \]

Step 5 - Calculation of Kinetic Energy After Collision

The kinetic energy after collision is zero because \(v=0\) (as established in Step 3). Hence, \[ KE_{final} = 0 \]

Step 6 - Calculation of Loss in Kinetic Energy

Subtract the final kinetic energy from the initial kinetic energy to find: \[ \text{Loss in kinetic energy} = KE_{initial} - KE_{final} = mu^2 - 0 = mu^2 \]

Step 7 - Calculation of Speed on Reaching the Ground

The combined mass moves to the ground under gravitational acceleration. Use the equation of motion for distance \(h\): \[ h = 0 \times t + \frac{1}{2} gt^2 \] to derive: \[ v = \sqrt{2gh} \]

Key Concepts

Time of Flight

The time of flight is the total time a projectile remains in the air. For the particle thrown upwards with initial speed \(u\), we start by determining how long it takes to reach its highest point. Using the equation of motion, we know that the time \( t \) to reach the height where it meets the particle thrown downwards can be found using \inline\frac{u}{g}\ w. Gravity slows down the particle at rate \(g = 9.8 \frac{m}{s^{2}}\). This gives us the equation: \inline t = \frac{u}{g}\. Understanding the time of flight helps determine when and where both particles meet, facilitating further calculations in the problem.

Conservation of Momentum

Conservation of momentum is a fundamental concept in collision scenarios. When the two particles collide, the total momentum before impact must equal the total momentum after impact. Prior to collision:

• The particle moving upwards has momentum: \(m \times u\).
• The particle moving downwards has momentum: \(-m \times u\).

At the point of impact, combining both gives zero net momentum: \(m \times u - m \times u = 0\). This principle simplifies evaluating the system's behavior following the collision, ensuring mass and velocity changes are accounted for accurately.

Kinetic Energy Loss

Kinetic energy loss in collisions tells us how much mechanical energy is converted into other forms of energy (like heat or sound). We first calculate initial kinetic energy with the formula: \(KE_{initial} = \frac{1}{2} m u^2 + \frac{1}{2} m u^2 = mu^2\). After impact, since the combined particle momentarily stops (velocity is zero), the final kinetic energy: \(KE_{final} = 0\). The difference between these gives the loss: \(mu^2 - 0 = mu^2\). Recognizing the energy lost underscores the significance of non-conservative forces at work during inelastic collisions.

Motion Under Gravity

Motion under gravity encompasses objects accelerating or decelerating solely due to gravity's pull. After the collision, the combined mass falls under gravitational acceleration from height \(h\) to the ground. Using the equation: \inline h = \frac{1}{2} g t^2\ derives the time taken to fall: \inline t = \frac{2h}{g}\. From here, the velocity of impact with the ground is found using \(v = \sqrt{2gh}\). Hence, \inline gh\ dictates the speed upon reaching the ground. This portrays gravity's predictable effect on objects, integral in calculating kinematic scenarios.