Question

a cricket ball moving horizontally with a velocity 12 m/s is brought to rest by a player in 0.1s if the cricket ball weighs .15kg calculate impulse of a force and avg force applied

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Answer 1

Answer :

• Impulse = - 180 Kg m/s

• Force = - 1800 N

Explanation :

• Initial Velocity,u = 12 m/s

• Mass of the ball,M = 15 Kg

• Time Taken,t = 0.1 s

Since the ball is brought to a rest,the following quantities would also bear negative sign

• Acceleration : indicating the decrease in velocity
• Force : Indicating the applied force is acting opposite to the motion of ball
• Impulse

Impulse

The change in momentum in a finite period of time due to a force applied on the particle

Mathematically,

$\boxed{ \boxed{ \sf{I = M(v - u)}}} \\ \\ \leadsto \: \sf{I = - Mu} \\ \\ \leadsto \: \sf{I = - (15)(12)} \\ \\ \leadsto \: \underline{ \boxed{ \sf{I = - 180 \:Kg \: {ms}^{ - 1} }}}$

Now,

$\sf{F = Ma} \\ \\ \longrightarrow \: \sf{F = M \times \dfrac{(v - u)}{t} } \\ \\ \longrightarrow \: \sf{F = 15 \times - \dfrac{12}{0.1} } \\ \\ \longrightarrow \: \underline{\boxed{\sf{F = - 1800 \: N}}}$

The same can be arrived at by using the relation,

$\sf{Force \: = \dfrac{Impulse}{Time} }$

Note

The average force here would be same as the force applied because the time interval is constricted,i.e. very small. In other words,the instantaneous force is equal to the average force

Answer 2

AnSwEr :

• Mass of ball (m) = 15 kg
• Initial Velocity (u) = 12 m/s
• Final Velocity (v) = 0 m/s
• Time (T) = 0.1 s

$\rule{200}{1}$

The formula for Impulse is :

$\Large \star {\boxed{\sf{\vec{I} \: = \: \big( mv \: - \: mu \big) }}} \\ \\ : \implies {\sf{\vec{I} \: = \: \big[ (15 \: \times \: 0) \: - \: (15 \: \times \: 12) \big] }} \\ \\ : \implies {\sf{\vec{I} \: = \: 0 \: - \: 180}} \\ \\ : \implies{\sf{\vec{I} \: = \: - \: 180 \: Kg \: ms^{-1}}}$

$\Large {\boxed{\boxed{\sf{Impulse \: = \: -180 \: Kg \: ms^{-1}}}}}$

Also,

$\Large \star {\boxed{\sf{\vec{I} \: = \: \vec{F_{avg}} \: \times \: t}}} \\ \\ : \implies {\sf{Force \: = \: \dfrac{Impulse}{time}}} \\ \\ : \implies {\sf{Force \: = \: \dfrac{-180}{0.1}}} \\ \\ : \implies {\sf{Force \: = \: - 1800 \: N}}$

$\Large {\boxed{\boxed{\sf{Force \: = \: - \: 1800 \: N}}}}$