The size of the impulse exerted on the ball is 5.4 kg m/s
Explanation:
The impulse exerted on an object is equal to the change in momentum of the object itself:
[tex]I=\Delta p = m(v-u)[/tex]
where:
I is the impulse
m is the mass of the object
v is the final velocity of the object
u is its initial velocity
In this problem, we have:
m = 0.12 kg is the mass of the ball
u = 45 m/s is the initial velocity of the ball
v = 0 is the final velocity (the ball is brought to rest)
Substituting into the equation, we find the impulse:
[tex]I=(0.12)(0-45)=-5.4 kg m/s[/tex]
where the negative sign means that the direction of the impulse is opposite to the initial direction of motion of the ball. Therefore, the size of the impulse is
[tex]I=5.4 kg m/s[/tex]
Learn more about impulse and change in momentum:
brainly.com/question/9484203
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