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A mass \(m_{1}\) moves with a great velocity. It strikes anobe mass \(m_{2}\) at rest in head-on collision. It comes back along is path with low speed after collision. Then (1) \(m_{1}>m_{2}\) (2) \(m_{1}

Short Answer

Expert verified
The correct relation is (2) \(m_1 < m_2\).

Step by step solution

01

Understanding the Question

We are given a scenario where a mass \(m_1\) collides with another mass \(m_2\) that is initially at rest. After the collision, mass \(m_1\) moves back along its initial path with a lower speed. We need to determine a relation between the masses of \(m_1\) and \(m_2\).
02

Apply Conservation of Momentum

The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. Before the collision, the initial momentum is \(m_1v_1 + m_2 \cdot 0 = m_1v_1\). After the collision, let the final velocities be \(v_1'\) for \(m_1\) and \(v_2'\) for \(m_2\). The conservation of momentum gives us \(m_1v_1 = m_1v_1' + m_2v_2'\).
03

Analyze the Elastic Collision Case

In an elastic collision, both momentum and kinetic energy are conserved. If \(m_1\) rebounds with a lower speed, \(v_1' < v_1\), it implies that some velocity has been transferred to \(m_2\). Generally, when a lighter object (\(m_1\)) strikes a heavier object (\(m_2\)), the lighter object rebounds with a decrease in speed. Thus, it suggests that \(m_1 < m_2\).
04

Consider the Inelastic Collision

If the collision is inelastic, only momentum is conserved but not kinetic energy. We are interpreting it such that \(m_1\) moves backward with a lesser speed. This scenario implies \(m_2\) is more significant to absorb the kinetic energy and send \(m_1\) backward. Thus \(m_1 < m_2\).
05

Conclusion

Both collision scenarios point towards \(m_1 < m_2\) as the most likely relationship given the description of the movement post-collision.

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Key Concepts

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

Elastic Collision
An elastic collision is a type of collision where not only momentum but also kinetic energy is conserved. This means the total kinetic energy of the system before and after the collision remains unchanged. In the scenario of an elastic collision, both objects involved don't stick together post-collision. This is an important characteristic that distinguishes elastic collisions from inelastic ones. For example, if mass \(m_1\) collides elastically with mass \(m_2\),
  • The momentum before collision: \(m_1v_1\)
  • The momentum after collision: \(m_1v_1' + m_2v_2'\)
  • Kinetic energy before collision: \(\frac{1}{2}m_1v_1^2\)
  • Kinetic energy after collision: \(\frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2\)
Both momentum and kinetic energy equations have to hold true simultaneously. In these kinds of collisions, typically when a lighter mass collides with a heavier mass, the lighter mass rebounds with reduced speed. This insight helps us deduce that \(m_1 < m_2\) when \(m_1\) rebounds at a decreased speed.
Inelastic Collision
An inelastic collision differs from an elastic one mainly because kinetic energy is not conserved, even though momentum is. This means while the total momentum remains constant, some kinetic energy is converted into other forms of energy, like heat or sound. In inelastic collisions, the colliding objects often stick together post-collision, although this is not always the case. Considering our scenario, after \(m_1\) strikes \(m_2\),
  • Momentum is conserved: \(m_1v_1 = m_1v_1' + m_2v_2'\)
  • Kinetic energy is not necessarily conserved
In these collisions, if \(m_1\) moves backward with reduced speed, likely the energy has been significantly absorbed or transferred to \(m_2\). Often, this supports the idea that \(m_1 < m_2\) as the larger mass is more capable of absorbing energy and sending the smaller mass back on its path.
Kinetic Energy Conservation
Kinetic energy conservation plays a crucial role in understanding collisions, especially elastic ones. Kinetic energy, the energy due to motion, is conserved in an elastic collision. It is defined mathematically as \(KE = \frac{1}{2}mv^2\). Analyzing conservation of kinetic energy:
  • Before collision: Total kinetic energy = \(\frac{1}{2}m_1v_1^2\)
  • After collision: Total kinetic energy = \(\frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2\)
Both sides of the equation should be equal in an elastic collision. In many real-world collisions, especially those dealing with macroscopic objects, there is often a degree of energy lost to the environment, making them inelastic. However, understanding the ideal condition of kinetic energy conservation allows students to better analyze and predict outcomes of collide scenarios.

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Most popular questions from this chapter

Two balls of masses \(m_{1}\) and \(m_{2}\) are moving towards each other with speeds \(u_{1}\) and \(u_{2}\), respectively. They collide head-on and their speeds are \(v_{1}\) and \(v_{2}\) after collision \(\left(m_{1}=8 \mathrm{~kg}, m_{2}=2 \mathrm{~kg}, u_{2}=3 \mathrm{~m} / \mathrm{s}\right)\) $$ \begin{array}{|l|l|} \hline \text { Column I } & {\text { Column II }} \\ \hline \begin{array}{l} \text { i. The speed } u_{1} \text { (in } \mathrm{m} / \mathrm{s} \text { ) so that both } \\ \text { balls move in same direction if } \\ \text { coefficient of restitution is } e=0.5 \end{array} & \text { a. } \frac{1}{14} \\ \hline \text { ii. The speed } u_{1} \text { (in } \mathrm{m} / \mathrm{s} \text { ) so that the } & \text { b. } \frac{1}{8} \\ \begin{array}{l} \text { maximum fraction of energy is } \\ \text { transformed to } m_{2} \text { (assume elastic } \\ \text { collision) } \end{array} & \\ \hline \begin{array}{l} \text { iii. Coefficient of restitution if } m_{2} \text { stops } \\ \text { after collision and } u_{1}=0.5 \mathrm{~m} / \mathrm{s} \end{array} & \text { c. } 2 \\ \hline \begin{array}{l} \text { iv. If collision is inelastic and } \\ u_{1}=3 \mathrm{~m} / \mathrm{s}, \text { the loss of kinetic energy } \\ \text { (in } \mathrm{J} \text { ) after collision may be } \end{array} & \text { d. } 4 \\ \hline \end{array} $$

In an elastic collision between two particles (1) the total kinetic energy of the system is always conserved (2) the kinetic energy of the system before collision is equal to the kinetic energy of the system after collision (3) the linear momentum of the system is conserved (4) the mechanical energy of the system before collision is equal to the mechanical energy of the system after collision

A steel ball of mass \(0.5 \mathrm{~kg}\) is tastened to a cord 20 cf long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes \(2.5 \mathrm{~kg}\) steel block initially at rest on a frictionless surface The collision is elastic. The speed of the block just after the collision will be (1) \(\frac{10}{3} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(\frac{20}{3} \mathrm{~ms}^{-1}\) (3) \(5 \mathrm{~ms}^{-1}\) (4) \(\frac{5}{3} \mathrm{~m} \mathrm{~s}^{-1}\)

A block of mass \(m\) starts from rest and slides down a frictionless semi- circular track from a height \(h\) as shown. When it reaches the lowest point of the track, it collides with a stationary piece of putty also having mass \(m\). If the block and the putty stick together and continue to slide, the maximum height that the block-putty system could reach is (1) \(h / 4\) (2) \(h / 2\) (3) \(h\) (4) independent of \(h\)

A block of mass ' \(m\) ' is hanging from a massless spring of spring constant \(k\). It is in equilibrium under the influence of gravitational force. Another particle of same mass ' \(m\) ' moving upwards with velocity \(u_{0}\) hits the block and sticks to it. For the subsequent motion, choose the incorrect statements: (1) Velocity of the combined mass must be maximum at natural length of the spring. (2) Velocity of the combined mass must be maximum at the new equilibrium position. (3) Velocity of the combined mass must be maximum at the instant particle hits the block. (4) Velocity of the combined mass must be maximum at a point lying between old equilibrium position and natural length.

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