Ask a Doubt
Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 11-science Answered

Define perfectly inelastic collision. Show that kinetic energy for perfectly inelastic collision is not conserved.
Asked by Sayoni Maiti | 08 Oct, 2014, 05:28: PM
answered-by-expert Expert Answer
 
begin mathsize 14px style The space collision space in space which space maximum space amount space of space kinetic space energy space has space been space lost space during space collision space is space called space perfectly space inelastic space collision. Let space straight m subscript 1 space be space the space mass space of space the space body space with space velocity space straight u subscript 1 space projected space towards space straight a space bdoy space of space mass space straight m subscript 2 space which space is space at space rest space left parenthesis straight u subscript 2 equals 0 right parenthesis After space the space collision comma space two space bodies space will space stick space together space and space with space moving space velocity space straight v. space As space the space total space linear space momentum space of space the space system space remain space constant comma therefore Pi equals straight P subscript straight f space space space space space space left parenthesis Pi equals Initial space momentum space and space straight P equals space Final space momentum right parenthesis rightwards double arrow straight m subscript 1 straight u subscript 1 plus straight m subscript 2 straight u subscript 2 equals left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v rightwards double arrow straight m subscript 1 straight u subscript 1 equals equals left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v space space space left parenthesis straight u subscript 2 equals 0 right parenthesis rightwards double arrow straight v equals fraction numerator straight m subscript 1 straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction To space calculate space loss space of space straight K. straight E. space in space this space collision Totla space straight K. straight E. space before space collision equals straight E subscript 1 equals 1 half straight m subscript 1 straight u subscript 1 squared Totla space straight K. straight E. space after space collision equals space straight E subscript 2 equals 1 half left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis straight v squared equals 1 half left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis open parentheses fraction numerator straight m subscript 1 straight u subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator straight m subscript 1 squared straight u subscript 1 squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction Loss space of space straight K. straight E. equals space straight E subscript 1 minus space space straight E subscript 2 equals space 1 half straight m subscript 1 straight u subscript 1 squared minus fraction numerator straight m subscript 1 squared straight u subscript 1 squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction equals fraction numerator straight m subscript 1 squared straight u subscript 1 squared plus straight m subscript 1 straight m subscript 2 straight u subscript 1 squared minus straight m subscript 1 squared straight u subscript 1 squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction space equals fraction numerator straight m subscript 1 straight m subscript 2 straight u subscript 1 squared over denominator 2 left parenthesis straight m subscript 1 plus straight m subscript 2 right parenthesis end fraction We space can space say space that space loss space of space straight K. straight E. space is space positive. space Hence space Kinetic space energy space for space perfectly space inelastic space collision space is space not space conserved. space space space space space space space space space space space space space space space space space end style
Answered by Priyanka Kumbhar | 09 Oct, 2014, 10:28: AM

Concept Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
CBSE 11-science - Physics
Can you tell an example about a body having 0 velocity and still accelerating?
Asked by veenasangeeth2020 | 30 Jan, 2024, 09:56: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
mention the expression for work done
Asked by abdulkhadirchajjo549 | 06 Dec, 2023, 06:55: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
During calculation of acceleration due to gravity above the surface of earth if we assume that there is an external force opposite to that of Earth's gravity then the cos angle will be 180 between the two forces which is negative(-1), if we put the value in the equation of dw = F.dx then the final value of potential energy is positive which is not possible.how and why??
Asked by anushkaverma.25707 | 24 Aug, 2023, 07:35: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
In the problem no. 34 of concepts of physics HC Verma-1 bit(a) , why is the acceleration of block B is half the acceleration of block A? And why is it that distance travelled by block B is half the distance travelled by block A?
question image
Asked by sakethrockzz007 | 04 Aug, 2023, 11:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
A small spherical object having density 2P is placed inside the liquid of density P and connected to an ideal string of length l whose other end is kept fixed so that the object can move in a vertical cirlcle, minimum speed which must be given at the bottom to just complete the vertical circle.
Asked by dhruvund8769 | 04 Feb, 2023, 11:41: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
For a liquid drop of radius r falling in air with terminal velocity power delivered by gravitational force is proportional to
Asked by gdeadshot85 | 12 Jan, 2023, 10:28: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
when we push the car,workdineyis 0,how
Asked by vaniimtiyaz630 | 12 May, 2022, 12:19: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
Two identical balls moving in opposite directions with speeds 20m/s and 25m/s undergo head on perfectly inelastic collision. The speed of combined mass after collision is 1. 20m/s 2. 22.5m/s 3. 25m/s 4. 2.5m/s
Asked by KRISHPATEL.soc | 22 Apr, 2022, 04:19: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
explain about conservation of energy with problem
Asked by bandarirohinipriya | 08 Feb, 2022, 09:46: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
CBSE 11-science - Physics
Derive expression for final velocity after perfectly inelastic collision  
Asked by unnatibhise08 | 22 Dec, 2021, 08:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt
Get Latest Study Material for Academic year 24-25 Click here
×