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.
Answer this Question
A. X = 36 m/s.
Y = 21 m/s.
V^2 = X^2 + Y^2 =
V^2 = 36^2 + 21^2 = 1737
V = 41.7 m/s.
Impulse = m*V = 0.23 * 41.7 = 9.59
B. tan A = Y/X = 21/36 = 0.58333
A = 30.3o = Direction.
To find the magnitude of the impulse delivered to the ball by the bat, we can use the principle of impulse-momentum. Impulse is defined as the change in momentum of an object and is calculated as the product of force and time. In this case, we can find the impulse by subtracting the final momentum of the ball from its initial momentum.
First, let's calculate the initial momentum of the ball:
Initial momentum = mass x initial velocity
= 0.230 kg x 36.0 m/s
= 8.28 kg·m/s
Next, let's find the final momentum of the ball:
Final momentum = mass x final velocity
= 0.230 kg x 21.0 m/s
= 4.83 kg·m/s
Now, we can calculate the change in momentum (impulse):
Impulse = Final momentum - Initial momentum
= 4.83 kg·m/s - 8.28 kg·m/s
= -3.45 kg·m/s (Note: the negative sign indicates that the impulse is in the opposite direction of the initial momentum).
Therefore, the magnitude of the impulse delivered to the ball by the bat is 3.45 kg·m/s.
For part B, since the initial direction of the ball is assumed to be the +x-axis direction, and the impulse delivered by the bat is in the opposite direction of the initial momentum, the direction of the impulse would be in the -x-axis direction.