with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? O 0.50 Joules O 3.75 Joules 2.5 Joules O 1.25 Joules O 5 Joules

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)...
icon
Related questions
icon
Concept explainers
Topic Video
Question
A 10.0 kg block is sliding along a frictionless table at 1.0 m/s. It collides with a stationary block
which also has a mass of 10.0 kg. Immediately after the collision, the blocks stick together and move
with a speed of 0.5 m/s. How much kinetic energy is lost in the collision?
0.50 Joules
O 3.75 Joules
2.5 Joules
O 1.25 Joules
O 5 Joules
Transcribed Image Text:A 10.0 kg block is sliding along a frictionless table at 1.0 m/s. It collides with a stationary block which also has a mass of 10.0 kg. Immediately after the collision, the blocks stick together and move with a speed of 0.5 m/s. How much kinetic energy is lost in the collision? 0.50 Joules O 3.75 Joules 2.5 Joules O 1.25 Joules O 5 Joules
Expert Solution
Step 1

Inelastic Collision

An inelastic collision is a type of collision where the kinetic energy is lost due to the internal friction of the particles. For a perfectly inelastic collision, the two colliding bodies stick together and move together as one unit. 

The above problem is an example of perfectly inelastic collision. Let two bodies of masses m1 and m2 be such that the first body is moving with an initial velocity v1i and the second body is at rest. 

The kinetic energy of the system before the collision

KEi=12m1v1i2+12m2.0=12m1v1i2

After the collision, the two bodies stick together and the two bodies move as a single unit. If vf is the velocity of the combined mass then the kinetic energy after the collision is

KEf=12m1+m2vf2

The loss of energy in this collision is

K=KEi-KEf=12m1v1i2-12m1+m2vf2

 

Step 2

Given in the question, the mass of the first block m1=10.0 kg

the mass of the second block m2=10.0 kg

 the initial velocity of the first block v1i=1 m/s.

the velocity of the combined system vf=0.5 m/s

 

Therefore the loss in kinetic energy

KE=12m1v1i2-12m1+m2vf2=12×10×12-12×10+10×0.52=2.5 J

steps

Step by step

Solved in 3 steps

Blurred answer
Knowledge Booster
Kinetic energy
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON