A 0.15 kg baseball is moving at 35 m/s toward home plate when it is hit with the bat. The ball is in contact with the bat for 0.0019 s, and leaves the bat at 25 m/s directly away from home plate. Determine the average force exerted by the bat on the ball.
avgforce(time)=.15(35+25)
solve for force
To find the average force exerted by the bat on the ball, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the rate of change of its momentum.
The momentum of an object is defined as the product of its mass (m) and velocity (v). So, the initial momentum of the baseball is given by:
Initial Momentum = mass * initial velocity = 0.15 kg * 35 m/s
The final momentum of the baseball can be calculated as:
Final Momentum = mass * final velocity = 0.15 kg * (-25 m/s) [note: the final velocity is negative because it is directed opposite to the initial velocity]
Since we know that the average force is equal to the change in momentum divided by the time interval:
Average Force = (Final Momentum - Initial Momentum) / Time
Now let's calculate the values and substitute them into the equation:
Initial Momentum = 0.15 kg * 35 m/s = 5.25 kg * m/s
Final Momentum = 0.15 kg * (-25 m/s) = -3.75 kg * m/s
Time = 0.0019 s
Average Force = (-3.75 kg * m/s - 5.25 kg * m/s) / 0.0019 s
Adding the initial momentum and the final momentum gives us -9 kg * m/s.
Average Force = (-9 kg * m/s) / 0.0019 s
Dividing -9 kg * m/s by 0.0019 s, we get:
Average Force = -4736.84 N
Therefore, the average force exerted by the bat on the ball is approximately -4736.84 Newtons. The negative sign indicates that the force is in the opposite direction of the initial momentum, which means the bat is applying an opposing force to the ball.