A block moving horizontally on a smooth surface with a speed of 40 m/s splits into parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is:-

Option 1 -

$\frac{2}{3}$

Option 2 -

$\frac{1}{3}$

Option 3 -

$\frac{1}{4}$

Option 4 -

$\frac{1}{8}$

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

Using conservation of linear momentum:

=> 40 × 3m = 60 × m + v × 2m

=> v = 30 m/s

$K E_{i} = \frac{1}{2} \times 3 m \times {(40)}^{2}$

$K E_{f} = \frac{1}{2} \times m \times {(60)}^{2} + \frac{1}{2} \times 2 m \times {(30)}^{2}$

$\Rightarrow \frac{K E_{f}}{K E_{i}} = \frac{54}{48}$

Fractional change in kinetic energy = 1 $- \frac{K E_{f}}{K E_{i}} = \frac{1}{8}$

Using conservation of linear momentum:=> 40 × 3m = 60 × m + v × 2m=> v = 30 m/s <math> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>i</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>3</mn> <mi>m</mi> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>4</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>      <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815997801"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mi>m</mi> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>6</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> <mi>m</mi> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>3</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>     <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815997802"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>f</mi> </mrow> </msub> </mrow> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>i</mi> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <mn>4</mn> </mrow> <mrow> <mn>4</mn> <mn>8</mn> </mrow> </mfrac> </mrow> </math> Fractional change in kinetic energy = 1 <math> <mrow> <mo>−</mo> <mfrac> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>f</mi> </mrow> </msub> </mrow> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>i</mi> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1815997804"> </o:OLEObject></xml><! [endif]->

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is:-

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question