If the ball started native rest, what impulse was applied to theball by the racket?

Express your answer making use of two far-ranging figures.

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The ideas used to resolve this difficulty are the momentum expression and readjust in momentum principle.

First calculation the momentum of the ball by making use of the inert formula. After ~ that, calculation the advertise by using readjust in inert principle.

The expression of inert is,

P=mvP = mvP=mv

Here, mmm is mass, and also vvv is velocity.

The impulse is equal to the adjust in momentum.

I=ΔP=Pf−Pi\beginarrayc\\I = \Delta P\\\\ = P_\rmf - P_\rmi\\\endarrayI=ΔP=Pf−Pi

Here, PfP_\rmfPf is final momentum, and PiP_\rmiPi is initial momentum.

Calculate the final momentum of the ball.

The expression of inert is,

P=mvP = mvP=mv

Here, mmm is mass, and also vvv is velocity.

Substitute 57g57\rm g57g for mmm , and also 45m/s45\rm m/s45m/s for vvv .

Pf=(57g)(45m/s)=(57g)(10−3kg1g)(45m/s)=(57×10−3kg)(45m/s)=2.6kg⋅m/s\beginarrayc\\P_\rmf = \left( 57\rm g \right)\left( 45\rm m/s \right)\\\\ = \left( 57\rm g \right)\left( \frac10^ - 3\rm kg1\rm g \right)\left( 45\rm m/s \right)\\\\ = \left( 57 \times 10^ - 3\rm kg \right)\left( 45\rm m/s \right)\\\\ = 2.6\rm kg \cdot \rmm/s\\\endarrayPf=(57g)(45m/s)=(57g)(1g10−3kg)(45m/s)=(57×10−3kg)(45m/s)=2.6kg⋅m/s

Calculate the in initial momentum.

Because the ball is initially in rest so the initial momentum of the round is zero.

Pi=0P_\rmi = 0Pi=0

Calculate the impulse applied to the ball by the rocket.

The advertise is same to the change in momentum.

I=ΔP=Pf−Pi\beginarrayc\\I = \Delta P\\\\ = P_\rmf - P_\rmi\\\endarrayI=ΔP=Pf−Pi

Here, PfP_\rmfPf is final momentum, and also PiP_\rmiPi is initial momentum.

Substitute 2.6kg⋅m/s2.6\rm kg \cdot \rmm/s2.6kg⋅m/s for PfP_\rmfPf , and 0kg⋅m/s0\;\rmkg \cdot \rmm/s0kg⋅m/s for PiP_\rmiPi .

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I=(2.6kg⋅m/s)−(0kg⋅m/s)=2.6kg⋅m/s\beginarrayc\\I = \left( 2.6\rm kg \cdot \rmm/s \right) - \left( 0\;\rmkg \cdot \rmm/s \right)\\\\ = 2.6\rm kg \cdot \rmm/s\\\endarrayI=(2.6kg⋅m/s)−(0kg⋅m/s)=2.6kg⋅m/s

Ans:The impulse applied to the round by the rocket 2.6kg⋅m/s2.6\rm kg \cdot \rmm/s2.6kg⋅m/s .