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Chapter 8: Problem 4

Two masses of 20 and 10 units. moving in the same direction at speeds of 16 and 12 units respectively collide and stick together. Find the velocity of the combined mass immediately afterwards.

Short Answer

Expert verified
The velocity of the combined mass immediately after the collision is 14.6 units.

Step by step solution

01

- Identify Given Values

Identify the masses and velocities of the objects involved. Let the masses be \( m_1 = 20 \) units and \( m_2 = 10 \) units, and their velocities be \( v_1 = 16 \) units and \( v_2 = 12 \) units respectively.
02

- Write Down the Conservation of Momentum

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The equation is given by: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f \]
03

- Substitute the Values into the Equation

Substitute the given values into the momentum equation. \[ 20 \times 16 + 10 \times 12 = (20 + 10)v_f \] \[ 320 + 120 = 30v_f \]
04

- Solve for the Final Velocity

Combine the terms on the left side and then solve for \( v_f \). \[ 440 = 30v_f \] Divide both sides by 30: \[ v_f = \frac{440}{30} \] \[ v_f = 14.\bar{6} \]

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Key Concepts

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

momentum
Momentum is a fundamental concept in physics. It quantifies the amount of motion an object has and is given by the product of its mass and velocity. The formula for momentum is: \[ p = mv \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In our exercise, we dealt with two objects, having masses of 20 and 10 units and velocities of 16 and 12 units respectively. By understanding momentum, you can predict how objects will move after interacting.
collision
In physics, a collision happens when two or more bodies exert forces on each other for a relatively short time. There are different types of collisions, such as elastic (where objects bounce off each other) and inelastic (where objects stick together). In this exercise, we are dealing with an inelastic collision because the two masses stick together after colliding. Understanding the type of collision helps in applying the right principles and equations to calculate the resulting motion.
velocity calculation
Calculating velocity is essential in solving many physics problems. In our exercise, we needed to find the velocity of the combined mass after the collision. Using the conservation of momentum formula: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f \] we can substitute the given values and solve for \( v_f \). After substituting, we get: \[ 20 \times 16 + 10 \times 12 = 30v_f \] \[ 440 = 30v_f \] By dividing both sides by 30, we find that the final velocity \( v_f \) is approximately 14.\bar{6} units. This shows how the velocities and masses of objects before a collision determine their combined velocity after the collision.
physics
Physics involves studying the natural laws that govern the behavior of matter and energy. Fundamental principles like the conservation of momentum allow us to understand and predict the outcomes of different scenarios. In this exercise, the law of conservation of momentum helped us determine the final velocity of two colliding masses. Knowing these principles not only helps with academic problems but also in practical applications in real life like car crashes, sports, and even space explorations.

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Most popular questions from this chapter

A ball falls vertically on to a horizontal plane and bounces. Find the impulse the ball exerts on the plane if: (a) the ball is initially \(2 \mathrm{~m}\) above the plane, (b) it rises after bouncing to a height \(1.2 \mathrm{~m}\), (c) the coefficient of restitution is \(\sqrt{\frac{2}{3}}\) (d) the mass of the ball is \(0.5 \mathrm{~kg}\).

A particle of mass \(4 \mathrm{~kg}\) is attached to one end \(\mathrm{X}\) of a light inextensible string which passes over a smooth light pulley and supports particies of masses \(2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) at the other end \(\mathrm{Y}\). The end \(\mathrm{X}\) is held in contact with a horizontal table at a depth \(6 \mathrm{~m}\) below the pulley, both portions of the string being vertical and the particles at \(\mathrm{Y}\) hanging freely. The system is released from rest. When \(\mathrm{Y}\) has descended a distance of \(2.5 \mathrm{~m}\), the particle of mass \(2 \mathrm{~kg}\) is disconnected and begins to fall freely. Calculate the greatest height reached by \(\mathrm{X}\) above the table and the momentum of the \(4 \mathrm{~kg}\) particle when it strikes the table. Take \(g\) to be \(9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Three particles A, B, C of masses \(m, 2 m, 3 m\) respectively lie at rest in that order in a straight line on a smooth horizontal table. The distance between consecutive particles is \(a .\) A slack light inelastic string of length \(2 a\) connects \(\mathrm{A}\) and \(\mathrm{B}\). An exactly similar slack string connects \(\mathrm{B}\) and \(\mathrm{C}\). If \(\mathrm{A}\) is projected in the direction CBA with speed \(V\), find the time which elapses before \(C\) begins to move. Find also the speed with which C begins to move. Show that the ratio of the impulsive tensions in \(\mathrm{BC}\) and \(\mathrm{AB}\) when \(\mathrm{C}\) is jerked into motion is \(3: 1 .\) Find the total loss of kinetic energy when C has started to move. (J.M.B.)

A golf ball, initially at rest, is dropped on to a horizontal surface and bounces directly up again with velocity \(v\). If the coefficient of restitution between the ball and the surface is \(e\), show that the ball will go on bouncing for a time \(\frac{2 v}{g(1-e)}\) after the first impact. \(\left\\{\right.\) You may assume \(\left.1+e+e^{2}+e^{3}+\ldots .=(1-e)^{-1}\right\\}\). If the golf ball is dropped from a height of \(19.62 \mathrm{~m}\) and comes to rest 12 seconds later, find the value of \(e\). (Take \(g\) as \(9.81 \mathrm{~ms}^{-2}\) ).

A pump raises water from a depth of \(10 \mathrm{~m}\) and discharges it horizontally through a pipe of \(0.1 \mathrm{~m}\) diameter at a velocity of \(8 \mathrm{~ms}^{-1} .\) Calculate the work done by the pump in one second. If the water impinges directly with the same velocity on a vertical wall, find the force exerted by the water on the wall if it is assumed that none of the water bounces back (Take \(g\) as \(9.81 \mathrm{~ms}^{-2}, \pi\) as \(3.142\) and the mass of \(1 \mathrm{~m}^{3}\) of water as \(1000 \mathrm{~kg}\) ).
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