A pitcher throws a 0.141-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 50 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat.
(a) What is the magnitude of the impulse delivered by the bat to the baseball?
(b) If the ball is in contact with the bat for 0.0046 s, what is the magnitude of the average force exerted by the bat on the ball?
(c) How does your answer to part (b) compare to the weight of the ball?
Fav/mg=
A) The magnitude of the impulse delivered by the bat to the baseball can be calculated using the principle of conservation of linear momentum. Since the ball is initially at rest and then moves with a speed of 50 m/s after being hit, the change in momentum is given by:
Δp = final momentum - initial momentum
= m * vf - m * vi
= m * (vf - vi)
Since the velocity is a vector quantity, we need to consider its direction. Since the ball is moving in the opposite direction after being hit, the velocity changes sign. Therefore, the magnitude of the impulse is given by:
Impulse = |Δp| = |m * (vf - vi)|
Substituting the given values:
Impulse = |0.141 kg * (50 m/s - (-42 m/s))|
= 0.141 kg * 92 m/s
= 12.252 kg·m/s
So, the magnitude of the impulse delivered by the bat to the baseball is 12.252 kg·m/s.
B) To find the magnitude of the average force exerted by the bat on the ball, we can use Newton's second law of motion, which states that force is equal to the rate of change of momentum.
Force = Δp / Δt
Substituting the given values:
Force = 0.141 kg * (50 m/s - (-42 m/s)) / 0.0046 s
= 0.141 kg * 92 m/s / 0.0046 s
= 2.82 kg·m/s^2 / 0.0046 s
= 613.04 N
So, the magnitude of the average force exerted by the bat on the ball is 613.04 N.
C) To compare the answer to part (b) to the weight of the ball, we can divide the magnitude of the average force by the weight of the ball.
Fav / mg = 613.04 N / (0.141 kg * 9.8 m/s^2)
= 613.04 N / 1.3818 N
= 443.97
So, the average force exerted by the bat on the ball is approximately 444 times the weight of the ball.
(a) To find the magnitude of the impulse delivered by the bat to the baseball, we need to find the change in momentum of the baseball.
The initial momentum of the baseball before it is hit by the bat is given by:
Initial momentum = mass * initial velocity = 0.141 kg * 42 m/s
The final momentum of the baseball after it is hit by the bat is given by:
Final momentum = mass * final velocity = 0.141 kg * (-50 m/s) [since the ball changes direction]
The change in momentum is therefore:
Change in momentum = Final momentum - Initial momentum = (0.141 kg * (-50 m/s)) - (0.141 kg * 42 m/s)
Calculate the value:
Change in momentum = -7.89 kg · m/s
The magnitude of the impulse delivered by the bat to the baseball is the absolute value of the change in momentum:
Magnitude of impulse = |Change in momentum| = |-7.89 kg · m/s|
Therefore, the magnitude of the impulse delivered by the bat to the baseball is 7.89 kg · m/s.
(b) To calculate the magnitude of the average force exerted by the bat on the ball, we use the equation:
Force = change in momentum / time
Plugging in the values:
Force = (-7.89 kg · m/s) / 0.0046 s
Calculate the value:
Force ≈ -1717.39 N
The magnitude of the average force exerted by the bat on the ball is the absolute value of the force:
Magnitude of force ≈ |-1717.39 N|
Therefore, the magnitude of the average force exerted by the bat on the ball is approximately 1717.39 N.
(c) To compare the answer to part (b) to the weight of the ball, we can calculate the weight of the ball using the equation:
Weight = mass * acceleration due to gravity = 0.141 kg * 9.8 m/s²
Calculate the value:
Weight ≈ 1.38 N
To compare the two values, we can calculate the ratio:
Force applied by the bat / Weight of the ball = |1717.39 N| / 1.38 N
Calculate the value:
Force applied by the bat / Weight of the ball ≈ 1245.97
Therefore, the answer to part (c) is approximately 1245.97.
To answer the questions, we need to apply the principles of impulse and momentum.
(a) The magnitude of the impulse delivered by the bat to the baseball can be calculated using the equation:
Impulse = Change in momentum
The momentum of an object is given by the equation:
Momentum = mass x velocity
In this case, the ball has a mass of 0.141 kg and a velocity of 50 m/s as it leaves the bat. Before it was hit by the bat, the ball had a velocity of 42 m/s. Therefore, the change in momentum is:
Change in momentum = (0.141 kg)(50 m/s) - (0.141 kg)(42 m/s)
Calculating this gives us the change in momentum delivered by the bat.
(b) To find the magnitude of the average force exerted by the bat on the ball, we can use the equation:
Average Force = Impulse / Time
We can plug in the value we previously calculated for impulse. The time of contact between the bat and the ball is given as 0.0046 s. By dividing impulse by time, you can find the average force exerted by the bat on the ball.
(c) To compare the answer in part (b) to the weight of the ball, we need to find the weight of the ball using its mass and the acceleration due to gravity. In this case, we use the equation:
Weight (Fg) = mass (m) x acceleration due to gravity (g)
The magnitude of the force exerted by the bat on the ball can be divided by the weight of the ball to compare the two values.
It crossed home plate horizontally ? - clever vertical plane curve ball :)
before:
Vx = 42
momentumx = Px = .141*42 = 5.922 kg m/s
after:
Vx = -50
momentum = Px = -.141*50 = -7.05 kg m/s
impulse = change of momentum = -7.05-5.922 = - 12.972 kg m/s
so 13 kg m/s (part A)
force = change of momentum/time
= 13/.0046 = 2820 Newtons (part B)
weight = m g = .141*9.81 = 1.38 Newtons
2820/1.38 = 2038 times the weight