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Chapter 2: Problem 3

A body of mass \(3 \mathrm{~kg}\) moving with a velocity of \(4 \mathrm{~m} / \mathrm{s}\) towards left collides head on with a body of mass \(4 \mathrm{~kg}\) moving in opposite direction with a velocity of \(3 \mathrm{~m} / \mathrm{s}\). After collision the two bodies stick together and move with a common velocity which is (1) zero (2) \(12 \mathrm{~m} / \mathrm{s}\) towards left (3) \(12 \mathrm{~m} / \mathrm{s}\) towards right (4) \(\frac{12}{7} \mathrm{~m} / \mathrm{s}\) towards left

Short Answer

Expert verified
The common velocity after collision is zero.

Step by step solution

01

Understanding the Problem

We have two bodies, one with mass \(3\, \text{kg}\) and velocity \(4\, \text{m/s}\) to the left, the other with mass \(4\, \text{kg}\) and velocity \(3\, \text{m/s}\) to the right. After collision, they stick together. We need to find their common velocity after the collision.
02

Identify the Formula for Conservation of Momentum

Since momentum is conserved in the collision, we can use the momentum conservation principle: \[ m_1v_1 + m_2v_2 = (m_1+m_2)V_f \] where \(m_1\) and \(m_2\) are the masses, \(v_1\) and \(v_2\) are the initial velocities, and \(V_f\) is the final velocity.
03

Assign Proper Signs to Velocities

As per the direction, let left be negative. So, \(v_1 = -4\, \text{m/s}\) and \(v_2 = 3\, \text{m/s}\) (since right is positive).
04

Substitute and Solve for Final Velocity

Substitute values in the conservation of momentum equation: \[ 3(-4) + 4(3) = (3 + 4)V_f \] Simplifying the expression: \[-12 + 12 = 7V_f \] \[0 = 7V_f \] \[V_f = 0 \]
05

Interpret the Result

The final velocity \(V_f\) is zero, indicating that after the collision, the combined mass comes to rest.

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Key Concepts

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

Conservation of Momentum
Momentum is a crucial concept in physics, particularly in understanding collisions. It is defined as the product of an object's mass and velocity. The conservation of momentum is a fundamental principle that states that in a closed system, free from external forces, the total momentum before a collision is equal to the total momentum after the collision. This principle is written mathematically as:
  • \( m_1v_1 + m_2v_2 = (m_1 + m_2)V_f \)
Here, \( m_1 \) and \( m_2 \) represent the masses of the two colliding bodies, \( v_1 \) and \( v_2 \) their velocities before collision, and \( V_f \) the common velocity after they collide and stick together. Understanding how to apply this equation helps in predicting post-collision velocities, crucial for situations involving inelastic collisions, like the problem we are analyzing.
Collision Physics
Collisions are interactions between two or more bodies with resulting impact forces. In collision physics, interactions are often classified as elastic or inelastic. In an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved as some kinetic energy is transformed into other forms of energy. In the given problem, the collision is inelastic. The two bodies stick together after the collision, indicative of a perfectly inelastic collision. Perfectly inelastic collisions are common in real-life scenarios, where tangled vehicles after an accident or clumped clay pieces post-impact exemplify this. Recognizing the type of collision is vital to applying the correct conservation principles for analysis.
Final Velocity Calculation
To find the final velocity after a collision, we use the conservation of momentum principle. In the problem, this involves substituting known values into the equation:1. Assign signs based on direction. Choose a direction as positive; here, right is positive
  • \( v_1 = -4\, \text{m/s} \) (left)
  • \( v_2 = 3\, \text{m/s} \) (right)
2. Substitute into the equation:
  • \( 3(-4) + 4(3) = (3 + 4)V_f \)
3. Simplify the equation:
  • \(-12 + 12 = 7V_f \)
4. Solve for \( V_f \):
  • 0 = 7V_f
  • \( V_f = 0 \)
This means that the final velocity is zero, indicating that both bodies come to rest after the collision. The exact steps and calculations ensure a clearer understanding of how conserved momentum determines the aftermath of collisions.

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Most popular questions from this chapter

A particle strikes a horizontal smooth floor with a velocity \(u\) making an angle \(\theta\) with the floor and rebounds with velocity \(v\) making an angle \(\phi\) with the floor. If the coefficient of restitution between the particle and the floor is \(e\), then (1) the impulse delivered by the floor to the body is $$m u(1+e) \sin \theta$$ (2) \(\tan \phi=e \tan \theta\) (3) \(v=u \sqrt{1-(1-e)^{2} \sin ^{2} \theta}\) (4) the ratio of final kinetic energy to the initial kinetic energy is \(\left(\cos ^{2} \theta+e^{2} \sin ^{2} \theta\right)\)

A stream of glass beads, each with a mass of 15 gram, comes out of a horizontal tube at a rate of 100 per second. The beads fall a distance of \(5 \mathrm{~m}\) to a balance pan and bounce back to their original height. How much mass (in \(\mathrm{kg}\) ) must be placed in the other pan of the balance to keep the pointer at zero?

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 ball is dropped from a height of \(45 \mathrm{~m}\) from the ground. The coefficient of restitution between the ball and the ground is \(2 / 3\). What is the distance travelled by the ball in 4th second of its motion? Assume negligible time is spent in rebounding. Le \(g=10 \mathrm{~m} \mathrm{~s}^{2}\) (1) \(5 \mathrm{~m}\) (2) \(20 \mathrm{~m}\) (3) \(15 \mathrm{~m}\) (4) \(10 \mathrm{~m}\)

A particle \(A\) suffers an oblique elastic collision with a particle \(B\) that is at rest initially. If their masses are the same, then after the collision (1) their KE may be equal (2) \(A\) continues to move in the original direction while \(B\) remains at rest (3) they will move in mutually perpendicular directions (4) \(A\) comes to rest and \(B\) starts moving in the direction of the original motion of \(A\)
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