A 145-g baseball travels at 35 m/s toward a baseball player's bat (the bat is the object of reference) and rebounds in the opposite direction at 40 m/s
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To solve this problem, you can use the principles of conservation of momentum and the law of reflection.
1. Conservation of momentum: According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Momentum is calculated by multiplying the mass of an object by its velocity.
Given:
Mass of baseball (m1) = 145 g = 0.145 kg
Initial velocity before collision (v1) = 35 m/s
Final velocity after collision (v2) = -40 m/s (since the direction is opposite)
2. Convert the mass to kilograms to ensure units are consistent: 0.145 kg.
3. Calculate the initial momentum before the collision:
Momentum before collision (p1) = m1 * v1
Substituting the given values:
p1 = 0.145 kg * 35 m/s = 5.075 kg·m/s
4. Apply the law of reflection: When an object rebounds off a surface, the magnitude of its velocity is unchanged, but its direction is reversed.
5. Calculate the final momentum after the collision:
Momentum after collision (p2) = m1 * v2
Substituting the given values:
p2 = 0.145 kg * (-40 m/s) = -5.8 kg·m/s
6. Apply the law of conservation of momentum: The total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum (p1) = Total final momentum (p2)
5.075 kg·m/s = -5.8 kg·m/s
7. Solve for the total mass of the system. Since the baseball and the bat combine to form the system, the total mass is the sum of their individual masses.
Total mass (m_total) = m1 + m_bat
We don't have the mass of the bat, but we can assume it has a much greater mass compared to the baseball. Therefore, we can neglect the mass of the bat in this calculation.
m_total ≈ m1 = 0.145 kg
8. Solve for the velocity of the bat after the collision:
p2 = m_total * v_bat
Substituting the values:
-5.8 kg·m/s = 0.145 kg * v_bat
Solving for v_bat:
v_bat = -5.8 kg·m/s ÷ 0.145 kg ≈ -40 m/s
Therefore, the baseball player's bat rebounds with a velocity of approximately -40 m/s, which is the same magnitude but opposite direction as the initial velocity of the baseball.