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 kinetic energy lost by the colliding body $A$ is

\frac{1}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8}{9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{4}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{8}{9}$

Body 'A' of mass $4 m$ moves with speed $u$ and body 'B' of mass $2 m$ is at rest.

So, initial linear momentum of system = $4 m u + 2 m \times 0 = 4 m u$

Let final velocity of body 'A' is $v_{1}$ and body 'B' is $v_{2}$

Then, final linear momentum of system = $4 m v_{1} + 2 m v_{2}$

As collision is elastic , $e = 1$

So, $v_{1} = \frac{(4 m - 2 m) \times u}{(4 m + 2 m)} + \frac{2 (2 m) \times 0}{(4 m + 2 m)} = \frac{2 m u}{6 m}$

= $\frac{u}{3}$

Similarly, $v_{2} = \frac{(2 m - 4 m) \times 0}{(4 m + 2 m)} + \frac{2 (4 m) u}{(4 m + 2 m)} = \frac{8 m u}{6 m} = \frac{4 u}{3}$

So, initial kinetic energy of body 'A' = $\frac{1}{2} (4 m) u ² = 2 m u ²$

Final kinetic energy of body 'A' = $\frac{1}{2} (4 m) \frac{u ²}{9} = \frac{2 m u ²}{9}$

So, change in kinetic energy = final kinetic energy - initial kinetic energy

= $\frac{2 m u ²}{9} - 2 m u ²$

= $2 m u ² [\frac{1}{9} - 1]$

= $- \frac{16 m u ²}{9}$ [ Here negative sign indicates that kinetic energy is lost after collision]

So, fractional loss in kinetic energy =

= $\frac{(16 m u ² / 9)}{(2 m u ²)}$

= $\frac{8}{9}$

Suggest Corrections

171

Similar questions

2

m

v

2 m

v

2 m

K . E .

[e = 1]

Join BYJU'S Learning Program