A 2.2-kg baseball is pitched to you at 21 m/s. You hit the ball back along the same path, and at the same speed. If the bat was in contact with the ball for 0.17 s, what is the magnitude of the average force the bat exerted on the ball?
Please show all work and provide any necessary equations. Thank you in advance for your help!
force*time=change momentum=mass(21-(-21))
solve for force
To find the magnitude of the average force the bat exerted on the ball, we can use the principle of impulse-momentum. The impulse applied to an object is equal to the change in momentum it experiences. Mathematically, impulse (J) is given by the equation:
J = Δp
Where Δp is the change in momentum of the object.
We can calculate the change in momentum (Δp) using the equation:
Δp = m * Δv
Where m is the mass of the ball, and Δv is the change in velocity of the ball.
Given:
Mass of the ball (m) = 2.2 kg
Initial velocity of the ball (v1) = 21 m/s
Final velocity of the ball (v2) = -21 m/s (since the ball is hit back along the same path)
The change in velocity (Δv) can be calculated as:
Δv = v2 - v1 = -21 - 21 = -42 m/s
Now, we can substitute the values into the equation:
Δp = m * Δv
= 2.2 kg * (-42 m/s)
= -92.4 kg.m/s
The negative sign indicates that the direction of momentum has reversed.
Since the average force (F_avg) is given by the equation:
F_avg = Δp / Δt
Where Δt is the time duration (0.17 s) for which the force is applied.
Now, substituting the values:
F_avg = -92.4 kg.m/s / 0.17 s
= -543.53 N
The magnitude of the average force the bat exerted on the ball is approximately 543.53 Newtons.