Body A of mass moving with speed collides with another body B of mass , at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

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.
Body A of mass $4 m$ moving with speed $u$ collides with another body B of mass $2 m$ , at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is [2019]

1 5/9

2 1/9

3 8/9

4 4/9

Ans.
(3)

According to conservation of momentum,

$4 m u_{1} = 4 m v_{1} + 2 m v_{2} \Rightarrow 2 (u_{1} - v_{1}) = v_{2}$ ...(i)

From conservation of energy,

$\frac{1}{2} (4 m) u_{1}^{2} = \frac{1}{2} (4 m) v_{1}^{2} + \frac{1}{2} (2 m) v_{2}^{2}$

$\Rightarrow 2 (u_{1}^{2} - v_{1}^{2}) = v_{2}^{2}$ ...(ii)

From (i) and (ii), $2 (u_{1}^{2} - v_{1}^{2}) = 4 {(u_{1} - v_{1})}^{2}$

$3 v_{1} = u_{1}$ ...(iii)

Now, fraction of loss in kinetic energy for mass $4 m,$

$\frac{Δ K}{K_{i}} = \frac{K_{i} - K_{f}}{K_{i}} = \frac{\frac{1}{2} (4 m) u_{1}^{2} - \frac{1}{2} (4 m) v_{1}^{2}}{\frac{1}{2} (4 m) u_{1}^{2}}$ ...(iv)

Substituting (iii) in (iv), we get $\frac{Δ K}{K_{i}} = \frac{8}{9}$

Want Us to Email you About Special Offers & Updates?