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Chapter 1: Problem 23

A particle of mass \(4 m\) is projected from the ground at some angle with horizontal. Its horizontal range is \(R\). At the highest point of its path it breaks into two pieces of masses \(m\) and \(3 m\), respectively, such that the smaller mass comes to rest. The larger mass finally falls at a distance \(x\) from the point of projection, where \(x\) is equal to (1) \(\frac{2 R}{3}\) (2) \(\frac{7 R}{6}\) (3) \(\frac{5 R}{4}\) (4) none of these

Short Answer

Expert verified
The distance x is equal to \( \frac{7R}{6} \), so the answer is (2).

Step by step solution

01

Analyze Initial Velocity

Let's denote the initial velocity of the particle as \( v_0 \) and the angle of projection as \( \theta \). The horizontal and vertical components of the initial velocity are \( v_0 \cos \theta \) and \( v_0 \sin \theta \), respectively.
02

Determine Time of Flight to Highest Point

Since the particle reaches the highest point at half the time of flight, the vertical velocity component becomes zero at this point. Hence, the time to reach the highest point is \( t = \frac{v_0 \sin \theta}{g} \), where \( g \) is the acceleration due to gravity.
03

Calculate Horizontal Range Before Break

The total horizontal range \( R \) is given by the equation \( R = \frac{v_0^2 \sin 2\theta}{g} \). This entire range is achieved if the particle does not break.
04

Analyze Motion After Break

At the highest point, the smaller mass \( m \) comes to rest, meaning it has zero velocity. The larger mass \( 3m \) moves with the original horizontal component of the velocity (\( v_0 \cos \theta \)) of the whole system.
05

Calculate Unchanged Horizontal Momentum

The horizontal momentum before breaking was \( 4m \cdot v_0 \cos \theta \). After breaking, the momentum is \( 3m \cdot v_0' = 4m \cdot v_0 \cos \theta \), where \( v_0' \) is the velocity of the larger mass. Solving, we get \( v_0' = \frac{4}{3} v_0 \cos \theta \).
06

Calculate Additional Distance Covered by Larger Mass

The larger mass continues horizontally with the velocity \( \frac{4}{3} v_0 \cos \theta \) for an additional time of \( \frac{v_0 \sin \theta}{g} \) until it hits the ground (same as time needed to fall). The additional distance covered is \( \left( \frac{4}{3} v_0 \cos \theta \right) \cdot \frac{v_0 \sin \theta}{g} = \frac{4}{3} \cdot \frac{v_0^2 \cos \theta \sin \theta}{g} = \frac{2R}{3} \) using the formula for range from Step 3.
07

Total Distance Covered by Larger Mass

The total horizontal distance from the point of projection, where the larger mass falls, is the sum of \( \frac{R}{2} \) (which is halfway point of the original path) and \( \frac{2R}{3} \). Therefore, the total distance \( x \) is \( \frac{R}{2} + \frac{2R}{3} = \frac{3R}{6} + \frac{4R}{6} = \frac{7R}{6} \).

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Key Concepts

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

Horizontal Range
The concept of horizontal range refers to how far a projectile travels along the horizontal axis. In projectile motion, the initial velocity and the angle of launch determine the horizontal range. For a projectile launched with an initial velocity \( v_0 \) at an angle \( \theta \), the formula for horizontal range \( R \) is:
\[ R = \frac{v_0^2 \sin 2\theta}{g} \]
Here, \( g \) stands for the acceleration due to gravity.
  • The angle \( \theta \) affects the path the projectile will follow in its motion.
  • The horizontal component \( v_0 \cos \theta \) influences how far horizontally the projectile will travel.
  • The vertical component \( v_0 \sin \theta \) affects how long the projectile stays in the air.
Understanding these components provides insight into different trajectories and affects how we calculate distances such as the horizontal range.
Conservation of Momentum
Conservation of momentum is a fundamental principle in physics stating that in the absence of external forces, the total momentum of a system remains constant. In our problem, a particle of mass \(4m\) breaks into two masses at its highest point.
  • Before the breakup, the total horizontal momentum is \( 4m \cdot v_0 \cos \theta \).
  • After the breakup, the momentum shifts to the remaining larger mass \( 3m \) which must still equal the initial momentum.
According to conservation of momentum:
\[ 4m \cdot v_0 \cos \theta = 3m \cdot v_0' \]
Solving for \( v_0' \), we find:
\[ v_0' = \frac{4}{3} v_0 \cos \theta \]
This ensures that momentum is accurately transferred, allowing us to calculate distances traveled after breakup.
Mass Breakup in Motion
The mass breakup in motion phenomenon involves a projectile breaking into parts while in flight. This occurs at the highest point of the projectile's path in our scenario.
  • The particle originally has a total mass of \(4m\).
  • It splits into two different masses: \(m\) and \(3m\).
  • The smaller mass comes to rest immediately, effectively stopping its horizontal motion.
  • The larger mass continues forward, propelled by its retained momentum.
The larger mass then behaves like a new projectile with different properties, moving horizontally over the remaining time of the fall. Understanding how the breakup alters a projectile's behavior helps in determining where the fragments will land, as calculated by summing horizontal travel before and after the breakup. The calculated total horizontal distance covered by the large mass is:
\[ x = \frac{7R}{6} \]
This breakup and redistribution information helps solve problems involving complex projectile motions and outcomes.

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Most popular questions from this chapter

A man stands at one end of the open truck which can run on frictionless horizontal rails. Initially, the man and the truck are at rest. Man now walks to the other end and stops. Then which of the following is true? (1) The truck moves opposite to direction of motion of the man even after the man ceases to walk. (2) The centre of mass of the man and the truck remains at the same point throughout the man's walk. (3) The kinetic energy of the man and the truck are exactly equal throughout the man's walk. (4) The truck does not move at all during the man's walk.

Two blocks \(m_{1}\) and \(m_{2}\) are pulled on a smooth horizontal surface, and are joined together with a spring of stiffness \(k\) as shown in figure. Suddenly, block \(m_{2}\) receives a horizontal velocity \(v_{0}\), then the maximum extension \(x_{m}\) in the spring is (1) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{m_{1}+m_{2}}}\) (2) \(v_{0} \sqrt{\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right) k}}\) (3) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{2\left(m_{1}+m_{2}\right) k}}\) (4) \(v_{0} \sqrt{\frac{m_{1} m_{2}}{\left(m_{1}+m_{2}\right) k}}\)

Consider a hollow sphere of mass \(M\) and radius \(R\) resting on a smooth surface. A smaller sphere of mass \(m\) and radius \(t\) is initially held at position \(P\) within the bigger sphere. If the small sphere is now released, it rolls down the inner surface of hollow sphere and finally stops at its bottom point \(Q\). Then: (1) horizontal displacement of the smaller sphere on smooth surface is \(\frac{M(R-r)}{M+m}\) (2) horizontal displacement of the bigger sphere on smooth surface is \(\frac{M(R+r)}{M+m}\) (3) horizontal displacement of the bigger sphere on smooth surface is \(\frac{m(R-r)}{M+m}\) (4) none of these

A body of mass \(1 \mathrm{~kg}\) initially at rest, explodes and breaks into three fragments of masses in the ratio \(1: 1: 3\). The two pieces of equal mass fly off perpendicular to each other with a speed of \(15 \mathrm{~ms}^{-1}\) each. What is the velocity of the heavier fragment? (1) \(10 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(5 \sqrt{3} \mathrm{~ms}^{-1}\) (3) \(10 \sqrt{3} \mathrm{~ms}^{-1}\) (4) \(5 \sqrt{2} \mathrm{~ms}^{-1}\)

Choose the correct statements from the following: (1) The general form of Newton's second law of motion is \(\vec{F}_{\mathrm{ext}}=m \vec{a}\) (2) A body can have energy and get no momentum. (3) A body having momentum must necessarily have kinetic energy. (4) The relative velocity of two bodies in a head-on elastic collision remains unchanged in magnitude and direction.
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