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A body of mass \(2 \mathrm{~kg}\) moving with a speed of \(3 \mathrm{~ms}^{-1}\) collides with a body of mass \(1 \mathrm{~kg}\) moving with a speed of \(4 \mathrm{~ms}^{-1}\). If the collision is one dimensional and completely inelastic, the speed of composite mass after the collision may be (a) \(\frac{3}{2} m s^{-1}\) (b) \(\frac{2}{3} \mathrm{~ms}^{-1}\) (c) \(4 \mathrm{~ms}^{-1}\) (d) \(\frac{10}{5} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The speed of the composite mass after the collision is \( \frac{10}{3} \) m/s.

Step by step solution

01

Identify the Initial Problem

We are given two bodies with masses of 2 kg and 1 kg moving with speeds of 3 m/s and 4 m/s respectively. A one-dimensional completely inelastic collision occurs, meaning the two bodies stick together post-collision, acting as a single mass.
02

Apply Conservation of Momentum

The principle of conservation of momentum states that in the absence of external forces, the total momentum before the collision is equal to the total momentum after the collision. Let's denote the speed of the composite mass after the collision by \( v \). Thus, using the formula: Total Initial Momentum = Total Final Momentum.
03

Calculate Initial Momentum

Calculate the total momentum before the collision: \( p_{initial} = (m_1 \times v_1) + (m_2 \times v_2) \). Here, \( m_1 = 2 \) kg, \( v_1 = 3 \) m/s; and \( m_2 = 1 \) kg, \( v_2 = 4 \) m/s. So, the initial momentum is \( p_{initial} = (2 \times 3) + (1 \times 4) = 6 + 4 = 10 \) kg m/s.
04

Set Up the Equation for Final Momentum

After the collision, both bodies move together, so the final momentum is \( (m_1 + m_2) \times v = (2 + 1) \times v = 3v \). According to conservation of momentum: \( 10 = 3v \).
05

Solve for the Final Speed \( v \)

Rearrange the equation \( 10 = 3v \) to solve for \( v \): \( v = \frac{10}{3} \).
06

Select the Correct Option

The solution for \( v \) is \( \frac{10}{3} \), which simplifies to approximately 3.33 m/s, not directly matching any of the multiple-choice options. Upon closer inspection, none of the options exactly matches 3.33 m/s, so there's an error or typo in the options provided.

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Key Concepts

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

Conservation of Momentum
The conservation of momentum is a key principle in physics that helps us understand how objects behave during and after collisions. It states that if there are no external forces acting on a system, the total momentum of the system before a collision will be equal to the total momentum after the collision.
This is especially useful in analyzing inelastic collisions, where two bodies collide and stick together, moving as a single object post-collision. When dealing with collisions, momentum is calculated using the formula:
  • Momentum, \( p = m \times v \)
  • Where \( m \) is the mass and \( v \) is the velocity.
In the original exercise, two bodies are involved in the collision. The first body has a mass of 2 kg and a velocity of 3 m/s, while the second has a mass of 1 kg and a velocity of 4 m/s before they collide.
This creates the following initial momentum:\[ p_{initial} = (2 \times 3) + (1 \times 4) = 10 \] (in kg m/s)Upon collision and sticking together, the final momentum is governed by the conservation principle:\[ p_{final} = (2 + 1) \times v = 3v \]By setting these equal, we can solve for the final velocity of the now combined mass.
Composite Mass Speed
When two bodies collide in a completely inelastic manner, they form a composite mass. This means they stick together and move with a common velocity. Determining the speed of this composite mass is a crucial part of solving collision problems.
In our exercise, we found that the total initial momentum was 10 kg m/s. After collision, the combined mass of the two bodies is 3 kg.To find the speed of the composite mass, we use the conservation of momentum:
  • Total Initial Momentum = Total Final Momentum
  • \[ 10 = 3v \]
By solving this equation, we divide the initial momentum by the total mass:\[ v = \frac{10}{3} \approx 3.33 \text{ m/s} \]This means that the final speed of the composite mass is 3.33 m/s. The options provided in the original exercise may have been incorrect as they do not match this value directly. But knowing how to calculate it gives us the correct insight into the problem.
One-Dimensional Collision
One-dimensional collisions are simpler than other types because the motion of all objects occurs along a single straight line. This type of collision allows us to easily use the conservation of momentum without considering angles or component vectors.
In one-dimensional collisions, calculations become straightforward as velocity and momentum are scalar quantities in these situations.
For our exercise, both bodies are moving along the same line which makes it a perfect example of a one-dimensional inelastic collision. During the collision, both masses adhere to each other perfectly, forming a single unit post-collision.
Such collisions are an excellent way to practice applying the conservation of momentum because they eliminate additional complexities like rotational motion or varying directions. Only the magnitude and the product of mass and velocity need to be considered. By focusing on these elements, students can gain a clearer understanding of basic collision dynamics and the fundamental law of conservation of momentum.

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Most popular questions from this chapter

A mass \(m\) moves with a velocity \(v\) and collides inelastically with another identical mass. After collision the lst mass moves with velocity \(\frac{v}{\sqrt{3}}\) in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision. [AIEEE 2005] (a) \(\underline{v}\) (b) \(\sqrt{3} v\) (c) \(\frac{2}{\sqrt{3}} v\). (d) \(\frac{v}{\sqrt{3}}\)

A ball hits the floor and rebounds after an inelastic collision. In this case choose the correct statement(s). (a) The momentum of the ball just after the collision is the same as that just before the collision (b) The mechanical energy of the ball remains the same in the collision (c) The total enerfy of the ball and carth is conserved [d) The total momentum of the ball and earth is conserved

When a meteorite burns in the atmosphere, then (a) the momentum conservation principle is applicable to the meteorite system (b) the energy of meteorite remains constant (c) the conservation principle of momentum is applicable to a system consisting of meteorites, earth and air molecules (d) the momentum of metearite remains constant

A bullet of mass \(50 \mathrm{~g}\) is fired from a gun of mass \(2 \mathrm{~kg}\). If the total kinetic energy produced is \(2050 \mathrm{~J}\), the kinetic energy of the bullet and the gun respectively are (a) \(200 j, 5\rfloor\) (b) \(2000 \mathrm{~J}, 50 \mathrm{~J}\) (c) \(5 \mathrm{~J}, 200 \mathrm{~J}\) (d) \(50 \mathrm{~J}, 2000 \mathrm{~J}\)

\((n-1)\) equal point masses each of mass \(m\) are placed at the vertices of a regular \(n\)-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the positive vector of centre of mass [NCERT Exemplat] (a) \(\frac{1}{n-1} a\) (b) \(-\frac{1}{(n-1)} a\) (c) \((n-1) a\) (d) \(-\left(\frac{n-1}{a}\right)\)

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