Answer:
The change of ball's momentum is 0.308 [tex]\frac{kg .m}{s}[/tex]
Explanation:
Given:
Mass of baseball [tex]m = 0.140[/tex] m
Downward velocity [tex]v' = -1.20[/tex] [tex]\frac{m}{s}[/tex]
Upward velocity [tex]v'' = 1[/tex] [tex]\frac{m}{s}[/tex]
Here we have to find change in momentum,
Momentum is given by,
[tex]P = mv[/tex]
Downward momentum is given by
[tex]P_{d} = mv'[/tex]
[tex]P_{d} = 0.140\times -1.20[/tex]
[tex]P_{d} = -0.168[/tex] [tex]\frac{kg .m}{s}[/tex]
Upward momentum is given by,
[tex]P_{u} = mv''[/tex]
[tex]P_{u} = 0.140 \times 1[/tex]
[tex]P_{u} = 0.140[/tex] [tex]\frac{kg.m}{s}[/tex]
So change of the ball's momentum is given by,
[tex]P_{u} - P_{d} = 0.140 - (- 0.168)[/tex]
[tex]P_{u} - P_{d} = 0.308[/tex] [tex]\frac{kg .m}{s}[/tex]
Therefore, the change of ball's momentum is 0.308 [tex]\frac{kg .m}{s}[/tex]
The change of the ball's momentum during the bounce will be "0.308 kg.m/s".
According to the question,
Baseball's mass, m = 0.140 m
Downward velocity, v' = -1.20 m/s
Upward velocity, v" = 1 m/s
We know the momentum,
→ P = mv then,
The downward momentum be:
→ [tex]P_d[/tex] = mv'
By putting the values,
= 0.140 × (-1.20)
= -0.168 kg.m/s
Now, the upward momentum be:
→ [tex]P_u[/tex] = mv"
= 0.140 × 1
= 0.140 kg.m/s
hence, The change will be:
→ [tex]P_u -P_d[/tex] = 0.140 - (-0.168)
= 0.308 kg.m/s
Thus the response above is right.
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