Question

a cricket ball of 100 g moving with a speed of 35 m/s is brought to rest by a player 0.04s find (a) the change in momentum of the ball (b) the force applied by the player​

Answer 1

Answer:

$\huge \mathcal \colorbox{pink}{Solution}$

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

mass of the ball (m) =100g or 0.1kginitial velocity (u)

$= 25 \frac{m}{s}$

time (t) = 0.025sec.

time (t) = 0.025sec.final velocity (v) = 0

time (t) = 0.025sec.final velocity (v) = 0Initial momentum = mu

time (t) = 0.025sec.final velocity (v) = 0Initial momentum = mu = 0.1×25

$\large\sf\ = 2.5 \frac{kgm}{s}$

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

Change in momentum = mv-mu

$\large\sf\ = 0 - 2.5 = - 2.5 \frac{kgm}{s}$

$\large \sf \ average \: force = \frac{changeinmomentum}{time} \\ = - \frac{2.5}{0.025} \\ = - 100 \frac{kgm}{s} = - 100 \: n$

The negative sign actually shows that the force was applied

The negative sign actually shows that the force was applied opposite to the direction of the motion of the ball.

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

Or

V= u + at

0= 25 + a (0.03)

a= - 2500/3 m/S2

F= ma= 0.15 kg

x - 2500/2

F= - 125 N

I am not sure which one is correct.