A baseball player pops a pitch straight up. The ball (mass 230g ) was traveling horizontally at 36.0m/s just before contact with the bat, and 21.0m/s just after contact.
Part A
Determine the magnitude of the impulse delivered to the ball by the bat.
Part B
Determine the direction of the impulse delivered to the ball by the bat. Assume the initial direction of the ball as +x-axis direction.
To determine the magnitude of the impulse delivered to the ball by the bat, we need to find the change in momentum of the ball.
The impulse delivered to an object is equal to the change in momentum of that object. Mathematically, impulse (J) is given by the equation:
J = Δp
where Δp represents the change in momentum.
Part A:
To find the magnitude of the impulse, we'll calculate the change in momentum of the ball. Momentum (p) is given by the equation:
p = m * v
where m is the mass of the ball and v is its velocity.
First, we'll calculate the momentum before contact:
p1 = m * v1
where m = 230 g (convert to kg: 230 g ÷ 1000 = 0.23 kg) and v1 = 36.0 m/s.
p1 = 0.23 kg * 36.0 m/s = 8.28 kg⋅m/s
Next, we'll calculate the momentum after contact:
p2 = m * v2
where v2 = 21.0 m/s.
p2 = 0.23 kg * 21.0 m/s = 4.83 kg⋅m/s
Now, we can find the change in momentum:
Δp = p2 - p1
Δp = 4.83 kg⋅m/s - 8.28 kg⋅m/s = -3.45 kg⋅m/s
The magnitude of the impulse, J, is the absolute value of the change in momentum:
J = |Δp| = |-3.45 kg⋅m/s| = 3.45 kg⋅m/s
Therefore, the magnitude of the impulse delivered to the ball by the bat is 3.45 kg⋅m/s.
Part B:
To determine the direction of the impulse, we need to consider the initial direction of the ball.
Given that the initial direction of the ball is the +x-axis direction, and the impulse has a magnitude but no direction, the impulse is in the same direction as the initial direction of the ball. Therefore, the direction of the impulse is in the +x-axis direction.