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
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.
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Tofind
(a) Change in momentum
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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.