Question

A cricket ball of mass 150g moving with a speed of 50m/s is brought to rest by a cricketer in 0.05s. Calculate:
(a) Change in momentum of ball
(b) the average force applied by the cricketer​

Answer 1

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

initial velocity (u) =25

s

m

time (t) = 0.025sec.

final velocity (v) = 0

Initial momentum = mu

= 0.1×25

= 2.5

s

kgm

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

Change in momentum = mv-mu

= 0 - 2.5 = -2.5

s

kgm

Average force =

time

Changeinmomentum

= -

0.025

(2.5)

= -100

s

kgm

= -100 N

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 be 100

Answer 2

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• A cricket ball of mass 150 g moving with a speed of $\sf {50ms}^{ - 1}$is brought to rest by a cricketer in 0.05 s.

$\huge \underline{ \underline {\rm{To \: find}} }$

(a) Change in momentum

$\huge \color{blue}{ \underline {\underline{ \cal \color{black}{S \sf olution}}}}$

Given;

m=150 g=$\sf \frac{150}{1000} kg = 0.15kg$

u=$\sf {50ms}^{ - 1}$

v=0

t=0.05 s

Initial momentum= mu= 0.15×50 =$\sf7.5 {kg \: ms}^{ - 1}$

Final momentum= mv=0.15×0= 0

Change in momentum = Final momentum - Initial momentum

$\sf \rightarrow0 - 7.5 = - 7.5 {ms}^{ - 1}$

(b) Average force applied by the cricketer

$\huge \color{blue}{ \underline {\underline{ \cal \color{black}{S \sf olution}}}}$

$\sf F=\frac {Change \: in \: momentum }{Time \: t}$

$\sf F= \frac{m(v - u)}{t} \rightarrow \frac{ - 7.5kg \: {ms}^{ - 1} }{0.05s} = - 150N$

Negative sign indicates that the force is aaplied in a direction opposite to the direction of motion of cricket ball.