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Chapter 4: Problem 512

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

Short Answer

Expert verified
The loss in kinetic energy due to the collision between the two balls is 60 J, denoted as (B) \(60\, \text{J}\).

Step by step solution

01

Calculate the initial velocity of the 2 kg ball in m/s

First, we need to convert the given velocity of the 2 kg ball from km/h to m/s. We will use the following conversion factor: 1 km/h = 1000 m / 3600 s \(v_1 = 36 \frac{\text{km}}{\text{h}} * \frac{1000 \text{m}}{3600 \text{s}}\) \(v_1 = 10\, \text{m/s}\) Now, the initial velocity for the 2 kg ball (v1) is 10 m/s.
02

Apply the conservation of linear momentum

The conservation of linear momentum states that the total initial momentum is equal to the total final momentum in a collision. For this exercise, the initial momentum (pi) can be calculated as: \(p_i = m_1v_1 + m_2v_2\) Since the 3 kg ball is stationary (v2 = 0), we have: \(p_i = m_1v_1 = 2\,\text{kg} * 10\,\text{m/s} = 20\, \text{kg m/s}\) After the collision, both balls move together with a combined mass of 5 kg (m1 + m2). Let vf be their final velocity: \(p_f = (m_1 + m_2)v_f\) Using the conservation of linear momentum: \(p_i = p_f\) We can now calculate the final velocity (vf): \(20\,\text{kg m/s} = 5\,\text{kg} * v_f\) \(v_f = \frac{20\,\text{kg m/s}}{5\,\text{kg}} = 4\,\text{m/s}\)
03

Calculate the initial and final kinetic energies

We can now calculate the initial kinetic energy (KEi) which is just the kinetic energy of the moving ball since the stationary ball has no kinetic energy: \(KE_i = \frac{1}{2}m_1v_1^2 = \frac{1}{2} * 2\,\text{kg} * (10\,\text{m/s})^2 = 100\,\text{J}\) And the final kinetic energy (KEf) is: \(KE_f = \frac{1}{2}(m_1+m_2)v_f^2 = \frac{1}{2} * 5\,\text{kg} * (4\,\text{m/s})^2 = 40\,\text{J}\)
04

Calculate the loss in kinetic energy

The loss in kinetic energy (ΔKE) can be calculated as: \(\Delta KE = KE_i - KE_f = 100\, \text{J} - 40\,\text{J} = 60\, \text{J}\) So, the loss in kinetic energy due to the collision is 60 J. Therefore, the correct answer is: (B) \(60\, \text{J}\)

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