The 0.060 kg tennis ball was traveling at 45 m/s (100 mph) and made a perfectly elastic collision with a wall and bounces back. If it is in contact with the wall for 0.15 s, what was the force the ball experienced from the wall?
To find the force the ball experienced from the wall, we can make use of Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of momentum. The momentum of an object is calculated by multiplying its mass by its velocity.
First, let's find the initial momentum of the tennis ball. Given that the mass of the ball is 0.060 kg and its initial velocity is 45 m/s, we can compute its momentum using the formula:
Initial momentum = mass × velocity
Initial momentum = 0.060 kg × 45 m/s
Now, let's calculate the final momentum of the ball after the collision. Since the collision is perfectly elastic, the ball bounces back with the same magnitude of velocity. Therefore, its final velocity is -45 m/s (opposite direction). Hence, we have:
Final momentum = mass × (final velocity)
Final momentum = 0.060 kg × (-45 m/s)
The change in momentum during the collision is the difference between the initial and final momentum. We can find this by subtracting the final momentum from the initial momentum:
Change in momentum = Final momentum - Initial momentum
Next, divide the change in momentum by the time of contact with the wall to determine the force experienced by the ball:
Force = (Change in momentum) / (Time of contact)
Force = (Final momentum - Initial momentum) / (Time of contact)
Plugging in the values into the above formula, we can calculate the force experienced by the ball.
acceleration = (45m/s)/(0.15s) = 300 m/s^2
F = ma