A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two spheres is

$90 °$

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

$0 °$

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

$45 °$

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

Different from $90 °$

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

Open in App

Solution

The correct option is D
Different from $90 °$

Explanation for the correct option $(s)$ :
$D :$ Different from $90 °$
An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision. Both momentum and kinetic energy are conserved in an elastic collision.
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Where,
$m_{1}, m_{2} =$ mass of object
$u_{1}, u_{2} =$ initial velocity
$v_{1}, v_{2} =$ final velocity
For perfectly elastic collision, the angle between path of two bodies must be $90 °$ .
An inelastic collision occurs when some amount of kinetic energy of a colliding object/system is lost.
$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v_{f}$
Where,
$m_{1}, m_{2} =$ mass of object
$u_{1}, u_{2} =$ initial velocity
$v_{f} =$ final velocity
Hence, for inelastic collision, angle should be different from $90 °$ .
Hence, option $' D'$ is correct.

Suggest Corrections

Similar questions

m

u

e

3 k g

{(\frac{1}{4})}^{t h}

3 k g

Join BYJU'S Learning Program