Question

A cricket ball of mass 200g is moving with speed of 36km/h. It is reflected back with the same speed. What is the impulse applied on it?

Answer 1

Answer:

Impulse = change in linear  momentum =mv1 -mv2 =m(v1-v2)

v=36 Km/h =36*5/18 m/s =10 m/s

and v2=-10 m/s (as the direction chosen +ve and -ve)

so,

I=200/1000 *(10- (-10))=0.2*20=4 kg m/s

Explanation:

Answer 2

Given that,

Mass of the ball(m)=200g =200/1000kg =0.2kg

Initial speed of the ball(v1)=36kmph =36*5/18 m/s =10m/s

Final speed of the ball(v2)=-36kmph =-36*5/18 m/s =-10m/s

To find,

The amount of impulse applied on the ball(I)

Solution,

Impulse(I)=mv2-mv1

Impulse(I)=m(v2-v1)

Impulse(I)=0.2(-10-10)

Impulse(I)=0.2*-20

Impulse(I)=-4kg m/s

Therefore,

The impulse applied on the ball of mass 200g with a speed 36kmph is -4kg m/s.