A 0.110 kg baseball, traveling 32.0 m/s, strikes the catcher's mitt, which recoils 14.0 cm in bringing the ball to rest. What was the average force (in newtons) applied by the ball to the mitt?
To find the average force applied by the ball to the mitt, we can use Newton's second law of motion which states that force is equal to the rate of change of momentum.
The momentum of an object is given by the product of its mass and velocity. In this case, the mass of the baseball is 0.110 kg and its initial velocity is 32.0 m/s. So the initial momentum of the baseball is:
Initial momentum = mass x velocity
= 0.110 kg x 32.0 m/s
= 3.52 kg·m/s
When the baseball comes to rest, its final velocity is 0 m/s. Therefore, the change in momentum is:
Change in momentum = Final momentum - Initial momentum
= 0 - 3.52 kg·m/s
= -3.52 kg·m/s
Since the catcher's mitt recoils in bringing the ball to rest, the momentum change is negative.
From the given information, we know that the recoil distance of the mitt is 14.0 cm. To find the force applied by the ball, we need to convert this distance into meters.
Recoil distance in meters = 14.0 cm / 100
= 0.14 m
Now, we can calculate the average force using the following formula:
Average force = Change in momentum / Recoil distance
Average force = (-3.52 kg·m/s) / (0.14 m)
= -25.14 N
The negative sign indicates that the force is in the opposite direction of the initial velocity of the baseball. Therefore, the average force applied by the ball to the mitt is 25.14 Newtons, in the opposite direction to the ball's initial motion.