Get Answers to all your Questions

header-bg qa

A ball of mass 3 Kg collides elastically with a wall with velocity 10 m/sec at an angle of 30 \degree with the wall . 

The change in momentum of ball is (in Kg m /s )

Option: 1

20


Option: 2

30


Option: 3

15


Option: 4

45


Answers (1)

best_answer

As we have learned

If target particle is massive in case of elastic collision -

\bar{v}_{1}=0

\bar{v}_{2}=-\bar{u}_{2}

 

- wherein

The lighter particle recoil with same speed and

the massive \left ( m_{2} > >m_{1} \right )remain practically at rest

u_{1}, v_{1}: initial and final speed of mass m1

u_{2},v_{2}: initial and final speed of mass m2

 

 

 

e = 1 = \frac{\vec {v_2 }- \vec {v_1}}{\vec {u_1}- \vec {u_2}}

 

1 = \frac{0 - \vec {v_1}}{v \cos \theta - 0}

v_1 = -v \cos \theta

 

Change in velocity of ball 

 \\\Delta v = v_1 - u _1 = -v \cos \theta - (+ v \cos \theta )\\ \\ \Delta v = - 2 v \cos \theta

Change in momentum = |M \Delta v |

= +2 mv \cos \theta = 2 \times 3 \times 10 \times 1/2 = 30 Kgm/s

 

 

 

Posted by

Gaurav

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks