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A body of mass \(2 \mathrm{~kg}\) moving with a velocity \(3 \mathrm{~m} / \mathrm{s}\) collides with a body of mass \(1 \mathrm{~kg}\) moving with a velocity of \(4 \mathrm{~m} / \mathrm{s}\) in opposite direction. If the collision is head on and completely inelastic, then (1) both particles move together with velocity \((2 / 3) \mathrm{m} / \mathrm{s}\) (2) the momentum of system is \(2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) throughout (3) the momentum of system is \(10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) (4) the loss of \(\mathrm{KE}\) of system is \((49 / 3) \mathrm{J}\)

Short Answer

Expert verified
Statements (1), (2), and (4) are correct; (3) is incorrect.

Step by step solution

01

Determine Initial Momentum of the System

The initial momentum is given by the formula \( p_{initial} = m_1 v_1 + m_2 v_2 \). Here, \( m_1 = 2 \) kg, \( v_1 = 3 \) m/s for the first body, and \( m_2 = 1 \) kg, \( v_2 = -4 \) m/s for the second body (negative because it's in the opposite direction). Thus, \( p_{initial} = 2 \times 3 + 1 \times (-4) = 6 - 4 = 2 \) kg m/s.
02

Apply Conservation of Momentum in Inelastic Collision

In an inelastic collision, the bodies stick together after the collision. The final momentum is thus the sum of the masses times their common velocity \( v_f \). Hence, \( p_{final} = (m_1 + m_2)v_f \). The conservation of momentum principle states that \( p_{initial} = p_{final} \), therefore, \( 2 = (2 + 1)v_f \) which means \( v_f = \frac{2}{3} \) m/s.
03

Verify Consistency of Momentum Statement

From Step 1, we found that the momentum was initially 2 kg m/s, meaning statement (2) is true. Statement (3) claims a momentum of 10 kg m/s, which is incorrect based on our calculations.
04

Calculate the Initial Kinetic Energy

The initial kinetic energy is given by the formula \( KE_{initial} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \). Substituting the known values, \( KE_{initial} = \frac{1}{2} \times 2 \times 3^2 + \frac{1}{2} \times 1 \times (-4)^2 = 9 + 8 = 17 \) J.
05

Calculate the Final Kinetic Energy

The kinetic energy after collision \( KE_{final} = \frac{1}{2} (m_1 + m_2)v_f^2 = \frac{1}{2} \times 3 \times \left(\frac{2}{3}\right)^2 = 2/3 \) J.
06

Calculate the Loss of Kinetic Energy

The loss in kinetic energy is \( KE_{loss} = KE_{initial} - KE_{final} = 17 - \frac{2}{3} = \frac{51}{3} - \frac{2}{3} = \frac{49}{3} \) J. This confirms statement (4) is correct.

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Key Concepts

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

Conservation of Momentum
When two objects collide, the principle of conservation of momentum is key to understanding the outcome. This principle states that, unless an external force acts on the system, the total momentum before and after the collision remains constant.

In our scenario, we have two bodies colliding head-on. One with mass 2 kg moving at 3 m/s and another with mass 1 kg moving at -4 m/s (negative because it's in the opposite direction).

The momentum before the collision is calculated as:
  • Momentum of the first body: \(2 \, \text{kg} \times 3 \, \text{m/s} = 6 \, \text{kg m/s}\)
  • Momentum of the second body: \(1 \, \text{kg} \times (-4) \, \text{m/s} = -4 \, \text{kg m/s}\)
  • Total initial momentum: \(6 - 4 = 2 \, \text{kg m/s}\)
After the inelastic collision, the bodies move together, and since momentum is conserved, their combined momentum stays the same at 2 kg m/s. Therefore, the velocity of the two bodies post-collision is \(\frac{2}{3} \, \text{m/s}\). This perfectly demonstrates the conservation of momentum.
Kinetic Energy Loss
In a collision, another critical element to examine is kinetic energy, which can sometimes change. Inelastic collisions, like the one described in our example, typically result in a loss of kinetic energy.

Before the collision, we calculated the kinetic energy:
  • First body: \(\frac{1}{2} \times 2 \, \text{kg} \times (3 \, \text{m/s})^2 = 9 \, \text{J}\)
  • Second body: \(\frac{1}{2} \times 1 \, \text{kg} \times (-4 \, \text{m/s})^2 = 8 \, \text{J}\)
  • Total initial kinetic energy: \(9 + 8 = 17 \, \text{J}\)
After the collision, as the objects move together, the final kinetic energy is determined by using their combined mass and velocity:
  • Combined mass: 3 kg
  • Combined velocity: \(\frac{2}{3} \, \text{m/s}\)
  • Final kinetic energy: \(\frac{1}{2} \times 3 \, \text{kg} \times \left(\frac{2}{3} \, \text{m/s}\right)^2 = \frac{2}{3} \, \text{J}\)
The loss in kinetic energy is therefore \(17 - \frac{2}{3} = \frac{49}{3} \, \text{J}\). This energy is "lost" as it’s transformed into other forms, like sound or thermal energy, a typical hallmark of inelastic collisions.
Head-On Collision
Head-on collisions occur when two objects hit each other directly along a line of travel. This type of collision emphasizes the exchange and conversion of momentum and energy.

In our exercise, we have a perfect head-on collision where one object moves towards the other at a velocity of -4 m/s, while the other moves towards it at 3 m/s. The velocities are directly opposed, ensuring that the interaction affects both objects directly and equally, which is characteristic of head-on collisions.

Upon impact, both kinetic energy and momentum come into play:
  • The principle of conservation of momentum dictates that the total momentum (momentum of body 1 plus momentum of body 2) remains constant before and after the collision.
  • However, because the collision is inelastic, kinetic energy isn't fully conserved and some is dissipated, typically as heat or sound.
  • The bodies stick together post-collision, moving as a single unit, another hallmark of an inelastic collision.
Head-on collisions provide a straightforward setup for studying these fundamental physics principles, clearly illustrating the balance and transfer of momentum and the inevitable loss of kinetic energy in inelastic interactions.

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Most popular questions from this chapter

A \(5 \mathrm{~kg}\) sphere is connected to a fixed point \(O\) by an inextensible string of length \(5 \mathrm{~m}\). The sphere is resting on a horizontal surface at a distance \(4 \mathrm{~m}\) from \(O\). Sphere is given a vertical velocity \(v_{0}\), and it moves freely till it reaches point \(P\), when string becomes taut. Find the maximum allowable velocity \(v_{0}\) (in \(\mathrm{m} / \mathrm{s}\) ), if impulse of tension in string is not to exceed \(6 \mathrm{~N} \mathrm{~s}\)

A ball of mass \(1 \mathrm{~kg}\) is dropped from a height of \(3.2 \mathrm{~m}\) on smooth inclined plane. The coefficient of restitution for the collision is \(e=1 / 2\). The ball's velocity become horizontal after the collision. (1) The angle \(\theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)\). (2) The speed of the ball after the collision \(=4 \sqrt{2} \mathrm{~ms}^{-1}\). (3) The total loss in kinetic energy during the collision is \(8 \mathrm{~J}\) (4) The ball hits the inclined plane again while travelling vertically downward.

A particle of mass \(m\) comes down on a smooth inclined plane from point \(B\) at a height of \(h\) from rest. The magnitude of change in momentum of the particle between position \(A\) (just before arriving on horizontal surface) and \(C\) (assuming the angle of inclination of the plane as \(\theta\) with respect to the horizontal) is (1) 0 (2) \(2 m \sqrt{(2 g h)} \sin \theta\) (3) \(2 m \sqrt{(2 g h)} \sin \left(\frac{\theta}{2}\right)\) (4) \(2 m \sqrt{(2 g h)}\)

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A body \(X\) with a momentum \(p\) collides with another identical stationary body \(Y\) one dimensionally. During the collision, gives an impulse \(J\) to body \(X\). Then coefficient of restitution is (1) \(\frac{2 J}{p}-1\) (2) \(\frac{J}{p}+1\) (3) \(\frac{J}{p}-1\) (4) \(\frac{J}{2 p}-1\)

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