The magnitude of the change in momentum of the stone is about 18.4 kg.m/s
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Let's recall Impulse formula as follows:
[tex]\boxed {I = \Sigma F \times t}[/tex]
where:
I = impulse on the object ( kg m/s )
∑F = net force acting on object ( kg m /s² = Newton )
t = elapsed time ( s )
Let us now tackle the problem!
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Given:
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
Asked:
magnitude of the change of momentum of the stone = Δp = ?
Solution:
Firstly, we will calculate the final speed of the ball as follows:
[tex]Ek = 70\% \ Ek_o[/tex]
[tex]\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )[/tex]
[tex]v^2 = 70 \% \ (v_o)^2[/tex]
[tex]v = - v_o \sqrt{70 \%}[/tex] → negative sign due to ball rebounds
[tex]v = - v_o \sqrt{0.7} \texttt{ m/s}[/tex]
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Next, we could find the magnitude of the change of momentum of the stone as follows:
[tex]\Delta p_{stone} = - \Delta p_{ball}[/tex]
[tex]\Delta p_{stone} = - [ mv - mv_o ][/tex]
[tex]\Delta p_{stone} = m[ v_o - v ][/tex]
[tex]\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ][/tex]
[tex]\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ][/tex]
[tex]\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ][/tex]
[tex]\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}[/tex]
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Grade: High School
Subject: Physics
Chapter: Dynamics
