Question

A ball of mass 30kg hits a rigid wall which is stationary with a speed 36km/hr and it renounces with same speed and makes an angle then the momentum acquired by the wall is 300 kg m/s and find the angle ..........
1.$\pi \div 3$
2.$\pi \div 4$
3.$\pi \div 6$
4.$2\pi \div 3$
[based on NEET]

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Answer 1

Answer:

Given:

Mass = 30 kg

Velocity = 36 km/hr

Momentum gained = 300 kg m/s

To find :

Angle of the path of ball with wall

Concept:

Since momentum is an vector Quantity, we can break it down into 2 perpendicular components . The component which is perpendicular to the wall changes it's direction . Hence the momentum change occurs.

Let the angle between the path of ball and wall be θ.

Conversion :

Velocity has to be converted from Km/hr to m/s ;

36 km/hr = 10 m/s

Diagram:

See the attached photo to understand better

Calculation:

So considering the momentum change perpendicular to wall , we get :

$mv \sin( \theta) - \{- mv \sin( \theta) \} = 300$

$= > 2mv \sin( \theta) = 300$

$= > 2 \times 30 \times 10 \sin( \theta) = 300$

$= > \sin( \theta) = \dfrac{1}{2}$

$= > \theta = \dfrac{\pi}{3}$

So final answer :

$\boxed{ \sf{ \red{ \bold{ \huge{ \theta = \dfrac{\pi}{3} }}}}}$