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The balls have different masses and speeds.
Rank the following from greatest to least.
a. momentum
b. the impulse needed to stop them
(a)
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To rank: The momentum of the balls from greatest to least.
Answer to Problem 20A
B, D, C, A.
Explanation of Solution
Given:
The mass and velocity of four different balls are as given in the figure 1.
Mass of the ball A is mA=1.0 kg
Velocity of the ball A is vA=9.0 m/s
Mass of the ball B is mB=1.2 kg
Velocity of the ball B is vB=8.5 m/s
Mass of the ball C is mC=0.8 kg
Velocity of the ball C is vC=12 m/s
Mass of the ball D is mD=5.0 kg
Velocity of the ball D is vD=2.0 m/s
Formula used:
The momentum (p) of an object of mass (m) moving with a velocity (v) is given by
p=mv (1)
Calculation:
Substituting the numerical values in equation (1) ,
The momentum of the ball A is,
pA=(1.0 kg)(9.0 m/s)=9.0 kg.m/s
The momentum of the ball B is,
pB=(1.2 kg)(8.5 m/s)=10.2 kg.m/s
The momentum of the ball C is,
pC=(0.8 kg)(12.0 m/s)=9.6 kg.m/s
The momentum of the ball D is,
pD=(5.0 kg)(2.0 m/s)=10.0 kg.m/s
Conclusion:
The rank of the balls having the momentum greatest to least is B, D, C, A.
(b)
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To rank: The impulse needed to stop the balls from greatest to least.
Answer to Problem 20A
B, D, C, A.
Explanation of Solution
Given:
Mass of the ball A is mA=1.0 kg
Initial velocity of the ball A is vAi=9.0 m/s
Final velocity of the ball A is vAf=0.0 m/s
Mass of the ball B is mB=1.2 kg
Initial velocity of the ball B is vBi=8.5 m/s
Final velocity of the ball B is vBf=0.0 m/s
Mass of the ball C is mC=0.8 kg
Initial velocity of the ball C is vCi=12 m/s
Final velocity of the ball C is vCf=0.0 m/s
Mass of the ball D is mD=5.0 kg
Initial velocity of the ball D is vDi=2.0 m/s
Final velocity of the ball D is vDf=0.0 m/s
Formula used:
Initial momentum of the ball is
pi=mvi
Final momentum of the ball is
pf=mvf
From the impulse - momentum theorem, the impulse needed to stop the ball can be written as
FΔt=pf−pi
Substituting for pf and pi ,
FΔt=mvf−mvi
⇒FΔt=m(vf−vi) (2)
Calculation:
Substituting the numerical values in equation (2) ,
The impulse needed to stop the ball A is,
FΔt=(1.0 kg)( 0.0 m/s−9.0 m/s)=− 9.0 kg.m/s
The impulse needed to stop the ball B is,
FΔt=(1.2 kg)( 0.0 m/s−8.5 m/s)=− 10.2 kg.m/s
The impulse needed to stop the ball C is,
FΔt=(0.8 kg)( 0.0 m/s−12.0 m/s)=− 9.6 kg.m/s
The impulse needed to stop the ball D is,
FΔt=(5.0 kg)( 0.0 m/s−2.0 m/s)=− 10.0 kg.m/s
Conclusion:
The rank of the impulse needed to stop the balls from greatest to least is B, D, C, A.
Chapter 8 Solutions
Conceptual Physics: The High School Physics Program
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