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Ball \(1,\) with a mass of 100 g and traveling at \(10 \mathrm{m} / \mathrm{s},\) collides head-on with ball \(2,\) which has a mass of \(300 \mathrm{g}\) and is initially at rest. What are the final velocities of each ball if the collision is (a) perfectly elastic? (b) perfectly inelastic?

Short Answer

Expert verified
In the elastic collision, the final velocities of ball 1 (v1) and ball 2 (v2) obtained from the equations (1) and (2) are the solutions. In the inelastic collision, both balls will move together with the same final velocity that is \(v1 = v2 = 2.5m/s\).

Step by step solution

01

Identify Given Variables and Unknowns

Ball 1 has initial mass \(m1 = 100g = 0.1kg\) and initial velocity \(u1 = 10m/s\). Ball 2 has initial mass \(m2 = 300g = 0.3kg\) and initial velocity \(u2 = 0m/s\) as it is at rest. The final velocities of ball 1 and ball 2 (v1 and v2 respectively) after the collision are the unknowns.
02

Find Final Velocities for Elastic Collision

In a perfectly elastic collision, both momentum and kinetic energy are conserved. Using these principles, we can establish two equations:\n Conservation of Momentum: \(m1*u1 + m2*u2 = m1*v1+ m2*v2\) which results in \(0.1*10 + 0.3*0 = 0.1*v1 + 0.3*v2; => 1=0.1*v1+0.3v2; => 10= v1+3v2(1)\) and Conservation of kinetic energy: \(\frac{1}{2}*m1*u1^2 + \frac{1}{2}*m2*u2^2 = \frac{1}{2}*m1*v1^2 + \frac{1}{2}*m2*v2^2\; => 0.5*0.1*10^2 + 0 = 0.5*0.1*v1^2 + 0.5*0.3*v2^2; => 5 = 0.05*v1^2 + 0.15*v2^2 (2)\). Solving these two equations simultaneously will give the values for \(v1\) and \(v2.\)
03

Find Final Velocities for Inelastic Collision

In a perfectly inelastic collision, only momentum is conserved. So, \(m1*u1 + m2*u2 = (m1+m2)v\), where \(v\) is the common final velocity. Substituting the known values \(0.1*10 + 0.3*0 = (0.1 + 0.3)v; => 1 = 0.4*v; => v = 2.5 m/s\). Here, \(v1 = v2 = 2.5 m/s\).

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Key Concepts

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

Elastic Collision
In an elastic collision, two bodies collide and rebound without lasting deformation or the generation of heat. During such a collision, two important physical quantities are conserved: momentum and kinetic energy.

Momentum conservation in this context means the total momentum before the collision equals the total momentum after the collision. Mathematically, this is expressed as:
\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]

Kinetic energy conservation means the total kinetic energy before the collision is equal to the total kinetic energy afterwards, can be written as:
\[ \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \]

To solve physics collision problems involving elastic collisions,

Step-by-Step Approach

can be particularly useful. Applying the conservation laws enables you to set up equations that contain the variables representing the initial and final velocities of the colliding objects (Step 2 in the provided solution). Solving these simultaneously will yield the post-collision velocities for both objects.
Inelastic Collision
An inelastic collision differs from an elastic collision because the colliding bodies may not rebound from each other; they might stick together or deform, leading to the generation of heat or sound. Despite these differences, one principle remains true: total momentum is still conserved.

However, unlike in elastic collisions, kinetic energy is not conserved in inelastic collisions. Some of it is transformed into other kinds of energy like internal energy, which is why after the collision, the objects may move with a common velocity. The momentum conservation equation is similar:
\[ m_1u_1 + m_2u_2 = (m_1 + m_2)v \]

After solving for the final common velocity, we conclude that both objects move together at the same speed after the collision (Step 3 in the provided solution).
Momentum Conservation
Momentum conservation is a fundamental concept in physics, applicable in both elastic and inelastic collisions. The law states that within a closed system, the total momentum must remain constant if no external forces act on it. Momentum, which is the product of an object's mass and velocity, can be thought of as the 'quantity of motion' an object has.

When solving problems pertaining to collisions, you will often use the momentum conservation principle to set up an equation representing the state before and after the event (Steps 2 and 3 in the provided solution). This principle is a valuable tool because it helps to simplify complex scenarios into solvable mathematical relationships.
Kinetic Energy Conservation
In physics, the conservation of kinetic energy is primarily discussed in the context of elastic collisions. Kinetic energy, the energy possessed by an object due to its motion, is only conserved in these types of collisions. The sum of the kinetic energies of all the colliding bodies before the impact will equal the sum of their kinetic energies afterward, assuming no energy is lost to sound, heat or deformation.

To apply this concept in calculations (like in Step 2 of the provided solution), you will often square the velocities in the equations that you set up to represent kinetic energy. For students tackling collision problems, it's important to recognize when this principle applies -- which is only in perfectly elastic collisions -- and to meticulously balance the energy on both sides of your equation.

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Most popular questions from this chapter

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You have been asked to design a "ballistic spring system" to measure the speed of bullets. A bullet of mass \(m\) is fired into a block of mass \(M .\) The block, with the embedded bullet, then slides across a frictionless table and collides with a horizon- tal spring whose spring constant is \(k\). The opposite end of the spring is anchored to a wall. The spring's maximum compression \(d\) is measured. a. Find an expression for the bullet's initial speed \(v_{\mathrm{B}}\) in terms of \(m, M, k,\) and \(d\) Hint: This is a two-part problem. The bullet's collision with the block is an inelastic collision. What quantity is conserved in an inelastic collision? Subsequently the block hits a spring on a frictionless surface. What quantity is conserved in this collision? b. What was the speed of a \(5.0 \mathrm{g}\) bullet if the block's mass is \(2.0 \mathrm{kg}\) and if the spring, with \(k=50 \mathrm{N} / \mathrm{m},\) was compressed by \(10 \mathrm{cm} ?\) c. What fraction of the bullet's initial kinetic energy is "lost"? Where did it go?

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