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Two putty-like masses \(m\) and \(2 m\) are travelling in the same straight line but in opposite directions, with speed \(u\), when they collide and unite. Find the magnitude and direction of the velocity of the combined masses and determine the loss in energy.

Short Answer

Expert verified
The final velocity of the combined masses is given by \(V = u/3\), and the direction is the same as that of the initial motion of the heavier mass. The loss in energy is given by the difference between the initial and final kinetic energies.

Step by step solution

01

Apply the Law of Conservation of Momentum

The law of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision. Let's denote the initial momentum of the lighter mass as \(-mu\), the initial momentum of the heavier mass as \(2mu\), and the final momentum of the combined mass as \(3mV\), where \(V\) is the final velocity. The equation for the conservation of momentum is: \[-mu + 2mu = 3mV.\] Solve this equation for \(V\) to find the final velocity of the combined masses.
02

Calculate the Initial and Final Kinetic Energies

The initial kinetic energy of the system is the sum of the kinetic energies of the two masses, which can be calculated using the formula \(K.E. = 1/2 mv^2\), where \(m\) is the mass and \(v\) is the velocity. Similarly, the final kinetic energy of the system is the kinetic energy of the combined mass, which has mass \(3m\) and velocity \(V\).
03

Determine the Loss in Energy

The loss of energy can be found by subtracting the final energy from the initial energy. If the final energy is less than the initial energy, then there has been a loss in energy. If the final energy is greater than the initial energy, then there has been a gain in energy. In this case, the loss of energy is found by subtracting the final kinetic energy from the initial kinetic energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Collision Mechanics
In physics, collision mechanics is a crucial concept that deals with how bodies interact when they collide. When two objects, like the putty-like masses in the exercise, come into contact, they exchange forces and velocities based on their masses and initial speeds. There are different types of collisions, including elastic and inelastic ones. In the given exercise, the collision is perfectly inelastic, which means that the two masses stick together after colliding. The law of conservation of momentum is key in solving such problems.
It tells us that the total momentum of a system is conserved in the absence of external forces.
  • The formula for momentum is given by the product of mass and velocity, so for an object with mass ( $m$) and velocity ( $v$), the momentum is ( $mv$).
  • In our exercise, we consider the initial and final momentum of both masses. For the system, we initially have momentum from both masses moving in opposite directions. The combined momentum after collision becomes a single entity moving with a new velocity.
This helps us to find important missing variables like the final velocity of united masses.
Kinetic Energy
Kinetic energy is a type of energy that a body possesses due to its motion. It depends on the mass of the object and the square of its velocity. This can be calculated using the formula \( KE = \frac{1}{2}mv^2 \). In the textbook exercise, you first need to find the initial kinetic energy of both masses before the collision.
  • The initial kinetic energy for each mass is calculated separately and then added together to get the system's total initial kinetic energy.
  • For the lighter mass ( \(m\)), the kinetic energy is \( \frac{1}{2}m(-u)^2 \), and for the heavier mass ( \(2m\)), it is \( \frac{1}{2}(2m)u^2 \).
After the collision, only one combined object exists, so the final kinetic energy is found using the combined mass and the velocity calculated previously. This gives us an understanding of how the initial and final energies compare and relate due to the change from independent objects into a single entity.
Energy Loss in Collisions
In inelastic collisions, such as the one in this exercise, some kinetic energy is lost. The energy is not destroyed; rather, it is transformed, often into heat or sound. The concept of energy loss helps us understand why the final kinetic energy of the system is lesser than the initial kinetic energy.
  • Initially, the sum of kinetic energies is calculated for each object. By finding the difference between the initial total kinetic energy and the final kinetic energy, we can quantify the energy lost in the collision.
  • The formula used for calculating the loss in energy is: \[\text{Energy Loss} = \text{Initial Kinetic Energy} - \text{Final Kinetic Energy}\]
This concept is important as it shows that while momentum is always conserved, kinetic energy isn't necessarily conserved in inelastic collisions. Understanding this helps in fields such as engineering, where designers need to minimize energy loss for efficiency.

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Most popular questions from this chapter

A ball is projected vertically upwards with a speed \(U\) from the floor of a room of height \(h\). It hits the ceiling and then returns to the floor, from which it rebounds, managing to just hit the ceiling a second time. Calculate the coefficient of restitution between the floor and the ball, and the ceiling and the ball, if they are equal.

A sphere \(A\) of mass \(m\), travelling with speed \(u\), is in collision with a similar stationary sphere \(B\) of mass \(2 m\). The direction of motion makes an angle of \(\theta\) with the line connecting their centres. If after the collision sphere \(A\) is travelling in a direction at right angles to its original motion, find the value of the coefficient of restitution between the two spheres.

Two particles \(A\) and \(B\), both of mass \(m\), are fastened to the ends of a piece of inextensible string and placed on a horizontal table. Particle \(A\) is then projected with speed \(v\) in a direction of \(45^{\circ}\) to the string, so that the string slackens. Calculate the impulse in the string when the string next tightens, and find the magnitude and direction of the velocity of \(A\) and \(B\) immediately afterwards.

For each of the following forces, calculate the impulse for the time \(t\) (in seconds) given and the final velocity if the impulse is applied to a body of mass \(3 \mathrm{~kg}\) that is moving with speed \(2 \mathrm{~ms}^{-1}\) in the direction of its motion: (a) \(F=25 \mathrm{~N} ; t=0\) to \(t=5\) (b) \(F=\sin ^{2}(2 \pi t) ; t=0\) to \(t=1\); (c) \(F= \begin{cases}5 t, & 0 \leqslant t \leqslant 5 \\ 25, & 5

A ball is dropped on to a horizontal plane. If the coefficient of restitution between the ball and the plane is \(e\), show that the average speed for each bounce decreases at a rate of \(e\) per bounce.

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