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In a typical Indian Bugghi (a luxury cart drawn by horses), a wooden plate is fixed on the rear on which one person can sit. A bugghi of mass \(200 \mathrm{~kg}\) is moving at a speed of \(10 \mathrm{~km} / \mathrm{h}\). As it overtakes a school boy walking at a speed of \(4 \mathrm{~km} / \mathrm{h}\), the boy sits on the wooden plate. If the mass of the boy is \(25 \mathrm{~kg}\), what will be the new velocity of the bugghi?

Short Answer

Expert verified
The new velocity of the bugghi with the boy is approximately 9.33 km/h.

Step by step solution

01

Understand the Initial Conditions

The bugghi is moving at a speed of 10 km/h with a mass of 200 kg. A boy who has a mass of 25 kg is walking at 4 km/h and will sit on the bugghi.
02

Apply Conservation of Momentum

According to the conservation of momentum, the total momentum before and after the boy sits on the bugghi should be equal. This gives us the equation: \( m_{b} v_{b} + m_{w} v_{w} = (m_{b} + m_{w}) v_f \), where \( m_{b} \) and \( v_{b} \) are the mass and velocity of the bugghi, \( m_{w} \) and \( v_{w} \) are the mass and velocity of the boy, and \( v_f \) is the final velocity of the bugghi and the boy together.
03

Set Up the Equation Using Given Values

Substitute the given values into the momentum equation: \( 200 \times \frac{10}{3.6} + 25 \times \frac{4}{3.6} = (200 + 25) v_f \). Note that we convert km/h to m/s by dividing by 3.6.
04

Solve for the Final Velocity

Calculate and solve the equation for the final velocity \( v_f \): \[ v_f = \frac{\left( \frac{200 \times 10}{3.6} + \frac{25 \times 4}{3.6} \right)}{225} \approx \frac{555.5556 + 27.7778}{225} \approx \frac{583.3334}{225} \approx 2.5915 \text{ m/s} \].
05

Convert the Final Velocity to km/h

Convert the final velocity back to km/h by multiplying by 3.6: \( v_f = 2.5915 \times 3.6 \approx 9.3294 \text{ km/h} \).

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Key Concepts

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

Inelastic Collision
In physics, an inelastic collision is a type of collision in which the involved objects stick together after impact. They do not bounce off each other. Hence, kinetic energy is not conserved, although momentum is. Understanding this is crucial for solving problems like the one with the Indian bugghi, where the boy and the bugghi become a single system after the collision.
  • During an inelastic collision, the combined mass of the system after the collision is the sum of the individual masses before the collision.
  • Despite the loss of kinetic energy, the momentum of the system remains constant.
  • This type of collision is common in everyday life, like when clay blobs merge or a person sits in a moving cart.
In our bugghi scenario, when the boy jumps onto the bugghi, it forms an inelastic collision because they combine and continue moving together at a new velocity.
Velocity Calculation
In physics problems like this one, calculating the resultant velocity after an event is essential. Once two masses combine in an inelastic collision, they have a new, shared velocity. This is what we call the final velocity.
  • The formula generally requires you to convert any velocities from km/h to m/s for uniformity and ease of calculation.
  • In the case of the bugghi, we take the initial speeds of both the bugghi and the boy, convert them from km/h to m/s, and use them in our formula.
  • After solving for the final velocity, it is often useful to convert it back to a more practical unit, like km/h, which was used initially for better context.
By understanding these steps, you can ensure accuracy in your calculations and a clearer understanding of the motion dynamics involved.
Momentum Equation
The core of this problem is understanding and applying the momentum equation. In momentum conservation problems, the total momentum before any interaction equals the total momentum after the interaction. The equation used is:\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \] where:
  • and are the masses of the two bodies.
  • and are their velocities before collision.
  • is the final velocity of the combined system.
It's vital to set up the equation correctly. Here, you factor in the direction of motion for all parties. Note how mass and velocity work together to form momentum as a vector quantity.
In our exercise, maintaining the correct units and substituting accurately is crucial to solve for the final velocity. This equation is a fundamental principle that applies across many scenarios in physics involving movement and interactions.

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Most popular questions from this chapter

A bullet of mass \(m\) moving at a speed \(v\) hits a ball of mass \(M\) kept at rest. A small part having mass \(m^{\prime}\) breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed \(v_{1}\) in the direction of the bullet. Find the velocity of the bullet after the collision.A bullet of mass \(m\) moving at a speed \(v\) hits a ball of mass \(M\) kept at rest. A small part having mass \(m^{\prime}\) breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed \(v_{1}\) in the direction of the bullet. Find the velocity of the bullet after the collision.

A uniform chain of mass \(M\) and length \(L\) is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor when a length \(x\) has reached the floor.

A block of mass \(2 \cdot 0 \mathrm{~kg}\) moving at \(2 \cdot 0 \mathrm{~m} / \mathrm{s}\) collides head on with another block of equal mass kept at rest. (a) Find the maximum possible loss in kinetic energy due to the collision. (b) If the actual loss in kinetic energy is half of this maximum, find the coefficient of restitution.

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