A body is dropped on ground from a height and after hitting the ground, it rebounds to a height . If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is . The value of is _

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A body is dropped on ground from a height $' h_{1}'$ and after hitting the ground, it rebounds to a height $' h_{2}'$ . If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is $\frac{x}{4}$ . The value of $x$ is _________. [2023]

Ans.
(375)

$Let V_{1} and V_{2} be velocities just before and just after hitting the floor.$

$\frac{V_{1}}{V_{2}} = 4 \Rightarrow V_{1} = 4 V_{2}$

$K E_{before} = \frac{1}{2} m V_{1}^{2}$

$K E_{after} = \frac{1}{2} m V_{2}^{2} = \frac{1}{2} \frac{m V_{1}^{2}}{16}$

$Δ K E = \frac{1}{2} m V_{1}^{2} = (\frac{1}{16} - 1) = \frac{- 15}{32} m V_{1}^{2}$

$% change = \frac{Δ K E}{K E_{before}} \times 100 % = \frac{- 15}{16} \times 100 % = \frac{- 375}{4} %$

Want Us to Email you About Special Offers & Updates?