JEE Main 20154 April
- Exam
- JEE Main
- Year
- 2015
- Date
- 4 April
- Subject
- Physics
- Chapter
- Center of Mass, Momentum and Collision
JEE Main 2015 · 4 April · Physics · Center of Mass, Momentum and Collision
A particle of mass m moving in the x direction with speed 2v is hit...
A particle of mass moving in the direction with speed is hit by another particle of mass moving in the direction with speed If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
The initial momentum of system is \overset{\to}{{P}_{i}} = m\left(2V\right)\overset{^}{i} + \left(2m\right)V\overset{^}{j}
According to question as
On perfectly inelastic collision the particles stick to each other.
By conservation of linear momentum
\overset{\to}{{P}_{f}} = \overset{\to}{{P}_{i}} ⇒ 3m\overset{\to}{{V}_{f}} = m2V\overset{^}{i} + 2mV\overset{^}{j}
⇒ \overset{\to}{{V}_{f}} = \dfrac{2V}{3}\left(\overset{^}{i} + \overset{^}{j}\right) ⇒ | {V}_{f} | = \dfrac{2\sqrt{2}}{3}V
loss in KE. of system
% Loss in KE
According to question as
On perfectly inelastic collision the particles stick to each other.
By conservation of linear momentum
\overset{\to}{{P}_{f}} = \overset{\to}{{P}_{i}} ⇒ 3m\overset{\to}{{V}_{f}} = m2V\overset{^}{i} + 2mV\overset{^}{j}
⇒ \overset{\to}{{V}_{f}} = \dfrac{2V}{3}\left(\overset{^}{i} + \overset{^}{j}\right) ⇒ | {V}_{f} | = \dfrac{2\sqrt{2}}{3}V
loss in KE. of system
% Loss in KE