Question

A cricket ball of mass 200gm moving with velocity 15m/s is brought to rest by a player in 0.05 secs. What is the impulse ????

Answer 1

Hey friend ......



Impulse = change in momentum

so,

$i = m(v - u)$



$0.2(0 - 15) = - 3ns$



Hope it helps you


$<marquee > prabhudutt$

Answer 2

mass of the ball (m) =100g or 0.1kg

initial velocity (u) =25 m/s

time (t) = 0.025sec.

final velocity (v) = 0

Initial momentum = mu

= 0.1kg*25m/s​

= 2.5 kgm/s

Final momentum = mv (which is equal to 0, since v is zero)

Change in momentum = mv-mu

= 0 - 2.5 = -2.5 kgm/s

Average force = Change in momentum / time

= -2.5 kgms-1 /0.025s

= -100 kgm/s2

= -100 N

The negative sign actually shows that the force was applied oppasite to the direction of the motion of the ball. So, the average force applied by the player will be 100 N.