A 135 g baseball traveling at a speed of 36 m/s hits a bat and moves back along its incoming trajectory with a speed of 47 m/s.
a) What is the impulse delivered to the ball by the bat? (Answer: 11.2kg * m/s)
b) If the duration of the bat-ball collision is 7 x 10^-4s, what is the average force exerted by the bat on the ball during this period? (Answer: 160000N)
Please show all work on how to get those answers.
a. Impulse = 0.135*(36+47) = 11.2
b. a = (V2-V1)/t
a = (47-(-36))/7*10^-4 = 118,571 m/s^2.
F = m*a = 0.135 * 118,571 = 16,007 N.
To find the impulse delivered to the ball by the bat, we can use the principle of impulse-momentum:
Impulse (J) = change in momentum
a) First, let's find the initial momentum (p1) of the ball before the collision. We can use the formula:
Momentum (p) = mass (m) * velocity (v)
Given:
Mass of the baseball (m) = 135 g = 0.135 kg
Initial velocity of the ball (v1) = 36 m/s
p1 = m * v1
= 0.135 kg * 36 m/s
= 4.86 kg·m/s
Next, let's find the final momentum (p2) of the ball after the collision. The ball moves back along its incoming trajectory, so its velocity is now in the opposite direction. Using the same formula:
Final velocity of the ball (v2) = -47 m/s
p2 = m * v2
= 0.135 kg * (-47 m/s)
= -6.345 kg·m/s
The impulse delivered to the ball by the bat is the change in momentum, which is given by:
Impulse (J) = p2 - p1
= (-6.345 kg·m/s) - (4.86 kg·m/s)
= -11.205 kg·m/s (rounded to 3 decimal places)
However, impulse is a vector quantity, so its direction matters. In this case, we are only interested in the magnitude of the impulse, so we remove the negative sign:
Impulse (J) = 11.205 kg·m/s (rounded to 3 decimal places)
Thus, the impulse delivered to the ball by the bat is 11.2 kg·m/s.
b) To find the average force exerted by the bat on the ball during the collision, we can use the formula:
Average force (F) = impulse (J) / time (t)
Given:
Duration of the bat-ball collision (t) = 7 x 10^-4 s
Using the impulse value we found in part a):
F = J / t
= 11.205 kg·m/s / (7 x 10^-4 s)
= 160000 N
The average force exerted by the bat on the ball during the collision is 160000 N.
To solve part (a), we need to use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it. We can calculate the impulse applied to the baseball by using the formula:
Impulse = Δp = m * Δv
where:
Δp = change in momentum
m = mass of the baseball
Δv = change in velocity
Given:
m = 135 g = 0.135 kg (convert grams to kilograms)
Initial velocity, u = 36 m/s
Final velocity, v = -47 m/s (since the ball moves back along its incoming trajectory)
To calculate Δv, we need to find the difference between the final velocity and the initial velocity:
Δv = v - u = -47 m/s - 36 m/s = -83 m/s
Now, we can substitute the values into the equation to find the impulse:
Impulse = Δp = m * Δv = 0.135 kg * (-83 m/s) = -11.205 kg*m/s
Since impulse is a vector quantity, the negative sign indicates the change in direction of the ball's momentum. To get the magnitude, we can take the absolute value:
Impulse = | -11.205 kg*m/s | = 11.205 kg*m/s ≈ 11.2 kg*m/s
Therefore, the impulse delivered to the ball by the bat is approximately 11.2 kg*m/s.
Moving on to part (b), the average force exerted by the bat on the ball can be found using the formula:
Force = Impulse / Time
where:
Impulse is the value we calculated in part (a)
Time = duration of the bat-ball collision = 7 x 10^-4 s
Substituting the known values:
Force = 11.2 kg*m/s / (7 x 10^-4 s)
To divide by a number in scientific notation, we can multiply by the reciprocal, so the calculation becomes:
Force ≈ 11.2 kg*m/s * (1 / (7 x 10^-4 s))
Force ≈ 11.2 kg*m/s * (1 / 0.0007 s)
Force ≈ 11.2 kg*m/s * 1428.57 s/kg
Force ≈ 160000 N
Therefore, the average force exerted by the bat on the ball during the collision period is approximately 160,000 N.