A 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1. The contact time is 0.01 seconds. What is the change in momentum of the baseball?
Answers
If a 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1, then the change in momentum of the baseball is 9.5 kgm/s.
Explanation:
Mass of the baseball = 0.10 kg
The initial velocity of the baseball, v₁ = 40 m/s
The final velocity after hitting straight to the pitcher, v₂ = - 55 m/s …. [since it is in the opposite direction]
We know that,
The change in momentum is given by,
∆p = p₂ – p₁ = mv₂ – mv₁ = m [v₂ – v₁]…… (i)
Where
p₁ = initial momentum of the baseball = mv₁
p₂ = final momentum of the baseball = mv₂
Now, substituting the given values in the formula (i), we get
∆p = m [v2 – v1] = 0.10 * [(-55) – (40)] = 0.10 * (-95) = - 9.5 kgm/s
Thus, the change in momentum of the baseball is [-9.5] kgm/s.
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