/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 102 A mass \(m=1 \mathrm{~kg}\) movi... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 102

A mass \(m=1 \mathrm{~kg}\) moving horizontally with velocity \(v_{0}=2 \mathrm{~m} / \mathrm{s}\) collides in elastically with a pendulum of same mass. Find the maximum change in potential energy (in Joule) of combined mass.

Short Answer

Expert verified
The maximum change in potential energy of the combined mass is \(\Delta U = 0.5\) Joules.

Step by step solution

01

Analyze the given information

We have a mass m = 1 kg moving horizontally with velocity v鈧� = 2 m/s which collides inelastically with a pendulum of the same mass. Since it is an inelastic collision, the two masses will stick together after the collision.
02

Apply conservation of momentum

Before the collision, only the first mass is moving, so the initial momentum (p_initial) is given by: \(p_{initial} = m \cdot v_{0}\) After the collision, both masses move together at a velocity v. Their total momentum (p_final) is: \(p_{final} = (m + m) \cdot v = 2m \cdot v\) According to the conservation of momentum, the initial momentum should be equal to the final momentum. So, we have: \(m \cdot v_{0} = 2m \cdot v\)
03

Determine the velocity after collision

Solve for the velocity (v) after the collision from the equation obtained in step 2: \(v = \frac{mv_{0}}{2m}\) Substituting the given values, we get: \(v = \frac{1kg \cdot 2m/s}{2 \cdot 1kg} = 1m/s\) So, the velocity of the masses after the collision is 1 m/s.
04

Apply conservation of mechanical energy

After the collision, the two masses will swing upward, and their mechanical energy will be conserved. Initially, the combined mass has kinetic energy (K) and no potential energy (U): \(E_{initial} = K + U = \frac{1}{2} \cdot 2m \cdot v^2\) At the highest point of the swing, the potential energy (U) is maximum, and the kinetic energy (K) becomes zero: \(E_{final} = K + U = mgh\) By the conservation of mechanical energy, we have: \(E_{initial} = E_{final}\)
05

Calculate the maximum height

Substitute the energy expressions from step 4 into the conservation of mechanical energy equation and solve for the maximum height (h): \(\frac{1}{2} \cdot 2m \cdot v^2 = mgh\) Divide both sides by mass (m) and solve for h: \(h = \frac{v^2}{2g}\) Using the value of v = 1 m/s and g = 9.81 m/s虏, we get: \(h = \frac{(1m/s)^2}{2 \cdot 9.81m/s^2} = 0.051m\) The maximum height reached by the combined mass is 0.051 m.
06

Find the maximum change in potential energy

Now that we have the maximum height (h), we can find the maximum change in potential energy (螖U): \(\Delta U = mgh\) Substitute the known values: \(\Delta U = 1kg \cdot 9.81m/s^2 \cdot 0.051m = 0.5 J\) The maximum change in potential energy of the combined mass is 0.5 Joules.

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Most popular questions from this chapter

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Potential energy (in joule) of a particle of mass \(1 \mathrm{~kg}\) moving in \(x-y\) plane is \(U=3 x+4 y\), here \(x\) and \(y\) are in meter. If at time \(t=0\), particle is at rest at point \(P(6 \mathrm{~m}\), \(4 \mathrm{~m}\) ). Then (A) acceleration of particle is \((3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}\). (B) time when it crosses \(y\)-axis is \(t=1 \mathrm{~s}\). (C) speed of particle when it crosses \(y\)-axis is \(10 \mathrm{~m} / \mathrm{s}\). (D) it crosses \(y\)-axis at \(y=-8 \mathrm{~m}\).
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