Two balls with masses m1 = 2kg and m2 = 4kg are moving with velocities of v1 = (5i – 3j+8k) m/s and v2 = (li +5j – 12k) m/s when they collide and stick together. What is: • The velocity of the balls after the collisions. • The type of collision described.
Two balls with masses m1 = 2kg and m2 = 4kg are moving with velocities of v1 = (5i – 3j+8k) m/s and v2 = (li +5j – 12k) m/s when they collide and stick together. What is: • The velocity of the balls after the collisions. • The type of collision described.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
Two balls with masses \( m_1 = 2 \, \text{kg} \) and \( m_2 = 4 \, \text{kg} \) are moving with velocities of
\[
\vec{v_1} = (5\hat{i} - 3\hat{j} + 8\hat{k}) \, \text{m/s}
\]
and
\[
\vec{v_2} = (1\hat{i} + 5\hat{j} - 12\hat{k}) \, \text{m/s}
\]
when they collide and stick together. What is:
- The velocity of the balls after the collision.
- The type of collision described.
**Solution:**
1. **Velocity After Collision:**
- When two objects stick together after a collision, it is an inelastic collision. The combined velocity can be found using the conservation of momentum:
\[
m_1\vec{v_1} + m_2\vec{v_2} = (m_1 + m_2)\vec{v_f}
\]
- Solve for \(\vec{v_f}\):
\[
\vec{v_f} = \frac{m_1\vec{v_1} + m_2\vec{v_2}}{m_1 + m_2}
\]
- Calculate:
\[
\vec{v_f} = \frac{2(5\hat{i} - 3\hat{j} + 8\hat{k}) + 4(1\hat{i} + 5\hat{j} - 12\hat{k})}{2 + 4}
= \frac{(10\hat{i} - 6\hat{j} + 16\hat{k}) + (4\hat{i} + 20\hat{j} - 48\hat{k})}{6}
\]
\[
= \frac{(14\hat{i} + 14\hat{j} - 32\hat{k})}{6}
= \left(\frac{14}{6}\hat{i} + \frac{14}{6}\hat{j} - \frac{32}{6}\hat{k}\right)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f7197f7-d5c6-491c-af54-a704a2aa86db%2F35bc368c-f186-4c16-82d1-eb7c38d64394%2F8q18akn_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Two balls with masses \( m_1 = 2 \, \text{kg} \) and \( m_2 = 4 \, \text{kg} \) are moving with velocities of
\[
\vec{v_1} = (5\hat{i} - 3\hat{j} + 8\hat{k}) \, \text{m/s}
\]
and
\[
\vec{v_2} = (1\hat{i} + 5\hat{j} - 12\hat{k}) \, \text{m/s}
\]
when they collide and stick together. What is:
- The velocity of the balls after the collision.
- The type of collision described.
**Solution:**
1. **Velocity After Collision:**
- When two objects stick together after a collision, it is an inelastic collision. The combined velocity can be found using the conservation of momentum:
\[
m_1\vec{v_1} + m_2\vec{v_2} = (m_1 + m_2)\vec{v_f}
\]
- Solve for \(\vec{v_f}\):
\[
\vec{v_f} = \frac{m_1\vec{v_1} + m_2\vec{v_2}}{m_1 + m_2}
\]
- Calculate:
\[
\vec{v_f} = \frac{2(5\hat{i} - 3\hat{j} + 8\hat{k}) + 4(1\hat{i} + 5\hat{j} - 12\hat{k})}{2 + 4}
= \frac{(10\hat{i} - 6\hat{j} + 16\hat{k}) + (4\hat{i} + 20\hat{j} - 48\hat{k})}{6}
\]
\[
= \frac{(14\hat{i} + 14\hat{j} - 32\hat{k})}{6}
= \left(\frac{14}{6}\hat{i} + \frac{14}{6}\hat{j} - \frac{32}{6}\hat{k}\right)
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