A 2.50 kg ball strikes a wall with a velocity of 10.0 m/s to the left. The ball bounces off with a velocity of 4.5 m/s to the right. If the ball is in contact with the wall for 0.10 s, what is the constant force exerted on the ball by the wall?
N to the right
To find the constant force exerted on the ball by the wall, we can use Newton's second law of motion. This law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's find the acceleration of the ball during its contact with the wall. We can use the equation:
acceleration = (final velocity - initial velocity) / time
Given:
Mass of the ball (m) = 2.50 kg
Initial velocity (u) = -10.0 m/s (to the left)
Final velocity (v) = 4.5 m/s (to the right)
Time of contact (t) = 0.10 s
Substituting the values into the equation, we have:
acceleration = (4.5 m/s - (-10.0 m/s)) / 0.10 s
acceleration = 14.5 m/s / 0.10 s
acceleration = 145 m/s^2
Now that we have the acceleration, we can calculate the force using Newton's second law:
force = mass * acceleration
Substituting the values, we get:
force = 2.50 kg * 145 m/s^2
force = 362.5 N
Since the ball is bouncing off the wall to the right, the constant force exerted on the ball by the wall is 362.5 N to the right.