Answer: [tex]-25.2 kg m/s^{2}[/tex]
Explanation:
The missing question her is: What is the change in momentum of the baseball?
According to the Impulse-Momentum Theorem, we have:
[tex]F \Delta t= \Delta p[/tex]
Where:
[tex]F[/tex] is the exerted force
[tex]\Delta t[/tex] is the time interval
[tex] \Delta p=p_{2}-p_{1}[/tex] is the change in momentum
So, in the case of the bat and the ball we have:
For the bat:
[tex]F_{bat} \Delta t=p_{fbat}-p_{obat}[/tex] (1)
For the ball:
[tex]F_{ball} \Delta t=p_{fball}-p_{oball}[/tex] (2)
Since [tex]\Delta t[/tex] is the same for both the bat and the ball, their impulses are equal in magnitude but in opposite direction according to Newton's 3rd law of motion. Hence:
[tex]F_{bat} = -F_{ball}[/tex] (3)
Then:
[tex]p_{fbat}-p_{obat}=-(p_{fball}-p_{oball})[/tex] (4)
Where:
[tex]p_{fbat}=m_{bat} V_{fbat}[/tex]
[tex]p_{obat}=m_{bat} V_{obat}[/tex]
Being [tex]m_{bat}=1.8 kg[/tex], [tex]V_{fbat}=28 m/s[/tex], [tex]V_{obat}=42 m/sm/s[/tex]
[tex]p_{fball}=m_{ball} V_{fball}[/tex]
[tex]p_{oball}=m_{ball} V_{oball}[/tex]
Being [tex]m_{ball}=0.2 kg[/tex]
Then (4) is rewritten as:
[tex]m_{bat}(V_{fbat} - V_{obat})=-m_{ball}(V_{fball}-V_{oball})[/tex] (5)
Since [tex]V_{fball}-V_{oball}=\Delta V_{ball}[/tex]:
[tex]m_{bat}(V_{fbat} - V_{obat})=-m_{ball}\Delta V_{ball}[/tex] (6)
Isolating [tex]\Delta V_{ball}[/tex]:
[tex]\Delta V_{ball}=\frac{m_{bat}(V_{fbat} - V_{obat})}{-m_{ball}}[/tex] (7)
[tex]\Delta V_{ball}=\frac{1.8 kg(28 m/s - 42 m/s)}{-0.2 kg}[/tex] (8)
[tex]\Delta V_{ball}=126 m/s[/tex] (9)
Now, the change in momentum of the baseball is:
[tex]\Delta p_{ball}=-m_{ball}\Delta V_{ball}[/tex] (10)
Substituting (9) in (10):
[tex]\Delta p_{ball}=-0.2 kg(126 m/s)[/tex] (11)
Finally:
[tex]\Delta p_{ball}=-25.2 kgm/s^{2}[/tex] This is the change in momentum of the baseball. Note it is negative because its initial velocity is greater than its final velocity
The change in momentum of the baseball is 25 kgm/s in the direction of the bat's motion
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of bat = m₁ = 1.8 kg
mass of baseball = m₂ = 0.2 kg
initial velocity of bat = v₁ = 42 m/s
final velocity of bat = v₁' = 28 m/s
Unknown:
the change in momentum of baseball = I_baseball = ?
Solution:
Law of conservation of momentum states that:
[tex]m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'[/tex]
[tex]m_2v_2 - m_2v_2' = m_1v_1' - m_1v_1[/tex]
[tex]- ( m_2v_2' - m_2v_2 ) = m_1v_1' - m_1v_1[/tex]
[tex]I_{baseball} = - I_{bat}[/tex]
[tex]I_{baseball} = - (m_1v_1' - m_1v_1)[/tex]
[tex]I_{baseball} = - (1.8(28) - 1.8(42))[/tex]
[tex]I_{baseball} = - (50.4 - 75.6)[/tex]
[tex]I_{baseball} = 25 \texttt{ kgm/s}[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Dynamics
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
