Physics, asked by ashley4454, 19 hours ago

A cricketer brought a ball of mass 200 g (moving with velocity 60ms - 1 ) to rest in time 0.06 sec. Find the - i) change in the momentum of ball ii) average force applied by the cricketer.​

Answers

Answered by DhiruRiluloveyou
1

Explanation:

Given

• A cricket ball of mass 150 g moving with a speed of \sf {50ms}^{ - 1}50ms

−1

is brought to rest by a cricketer in 0.05 s.

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

Tofind

(a) Change in momentum

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

Solution

Given;

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

1000

150

kg=0.15kg

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

−1

v=0

t=0.05 s

Initial momentum= mu= 0.15×50 =\sf7.5 {kg \: ms}^{ - 1}7.5kgms

−1

Final momentum= mv=0.15×0= 0

Change in momentum = Final momentum - Initial momentum

\sf \rightarrow0 - 7.5 = - 7.5 {ms}^{ - 1}→0−7.5=−7.5ms

−1

(b) Average force applied by the cricketer

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Solution

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

Timet

Changeinmomentum

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

t

m(v−u)

0.05s

−7.5kgms

−1

=−150N

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

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