a baseball has a mass of 0.145kg. the ball is crossing the plate horizontally at 40 m/s when the batter hits the ball. The final velocity of the ball is 50 m/s stright up. what is the impulse that the batter gives the ball?
The magnitude of the velocity change is V = sqrt(40^2 + 50^2)= ___
The impulse is M*V = ___ kg m/s
Dudeeeee,
I couldn't figure this one out either. AND I WROTE THE TEST!
To find the impulse that the batter gives the ball, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it.
The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v):
p = m * v
In this case, we have the initial and final velocities of the ball, but we need to find the change in velocity. The change in velocity (Δv) can be calculated by subtracting the initial velocity (vi) from the final velocity (vf):
Δv = vf - vi
Now that we have the change in velocity, we can calculate the impulse (J) applied to the ball using the formula:
J = m * Δv
Given that the mass of the baseball (m) is 0.145 kg, the initial velocity (vi) is 40 m/s, and the final velocity (vf) is 50 m/s, we can plug these values into the equation to find the impulse:
Δv = 50 m/s - 40 m/s = 10 m/s
J = 0.145 kg * 10 m/s = 1.45 kg⋅m/s
Therefore, the batter gives the ball an impulse of 1.45 kg⋅m/s