A 0.145-kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to a height of 43 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] during contact. [Let the positive axis lie along the line from the batter to the pitcher, with the batter at the origin.] Please give magnitude and direction.
im so lost that its not even funny... idk what im doing and it wouldve been better if this person wouldve used numbers instead of letters =(
To calculate the average force exerted by the bat on the ball, we can use the principle of conservation of momentum. The change in momentum of an object is equal to the impulse applied to it.
The impulse acting on the baseball is equal to the average force multiplied by the contact time. Therefore, we need to find the change in momentum of the baseball when it is struck and popped up.
The initial momentum of the baseball (before being struck) can be calculated as:
Initial Momentum = mass * initial velocity
m1 * v1 = (0.145 kg) * (27 m/s) = 3.915 kg*m/s (taking upward as the positive direction)
The final momentum of the baseball (after being struck) can be calculated as:
Final Momentum = mass * final velocity
Since the baseball is popped straight up, the final velocity is zero:
m2 * v2 = (0.145 kg) * 0 = 0 kg*m/s
The change in momentum is given by:
Δp = p2 - p1
Δp = 0 - 3.915 kg*m/s = -3.915 kg*m/s
To calculate the average force, we can use the equation:
Average Force = Impulse / Contact Time
Average Force = Δp / Δt
Average Force = (-3.915 kg*m/s) / (2.35 ms) = -1668.085 kg*m/s^2
The magnitude of the average force is approximately 1668.085 N.
The direction of the force is negative, indicating that it is applied in the opposite direction of the positive axis (from the batter to the pitcher).
To calculate the average force exerted by the bat on the ball during contact, you can use Newton's second law of motion, which states that force equals mass multiplied by acceleration.
First, let's find the initial velocity of the baseball vertically when it is popped straight up after being struck by the bat. Since the baseball is pitched horizontally, its initial vertical velocity is 0 m/s.
The final velocity of the baseball when it reaches its maximum height is also 0 m/s because at the highest point, it stops momentarily before starting to fall back down.
Using the equation for uniformly accelerated motion in the vertical direction:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s)
u = initial velocity (0 m/s)
a = acceleration
s = displacement (43 m)
Rearranging the equation, we can solve for acceleration:
a = (v^2 - u^2) / (2s)
a = (0^2 - 0^2) / (2 * 43)
a = 0 / 86
a = 0 m/s^2
Since the acceleration is zero, this means that the baseball traveled at a constant speed in the vertical direction while reaching its maximum height.
Now, let's calculate the initial horizontal velocity of the baseball before it was struck by the bat. The given information tells us that the baseball was pitched horizontally at 27 m/s.
To calculate the average force, we need to find the change in momentum of the baseball. Momentum is the product of mass and velocity.
The change in momentum is equal to the final momentum minus the initial momentum. Since the final velocity in the horizontal direction is also 27 m/s (assuming no other external forces act on the ball after contact), the change in momentum is:
Δp = m * vf - m * vi
where:
m = mass of the baseball (0.145 kg)
vf = final velocity in the horizontal direction (27 m/s)
vi = initial velocity in the horizontal direction (27 m/s)
Δp = 0.145 kg * 27 m/s - 0.145 kg * 27 m/s
Δp = 0 kg·m/s
The change in momentum is zero because the initial and final velocities are the same.
According to Newton's third law of motion, the baseball exerts an equal and opposite force on the bat. Therefore, the average force exerted by the bat on the ball during contact is also zero.
In summary, the magnitude of the average force exerted by the bat on the ball during contact is 0 N, and the direction of the force is zero or no force exerted.
assuming the 27 was really horizontal, then the change in velocity has to be determined.
To get 43m high (that is pretty high)...
1/2 m vi^2=mg(43)
vi=sqrt (86*9.8)
change momentum=massball*deltaV
ok,to get deltaV, deltaV=sqrt(86*9.8) j +27i)
then force= changemometum/time
you get direction from the velocity coordinates.