A 3.3 kg ball strikes a wall with a velocity of
8.4 m/s to the left. The ball bounces off with
a velocity of 6.8 m/s to the right.
If the ball is in contact with the wall for
0.27 s, what is the constant force exerted on
the ball by the wall?
Answer in units of N
To find the constant force exerted on the ball by the wall, we need to use Newton's second law of motion. Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, we can calculate the change in velocity (acceleration) of the ball by subtracting the initial velocity from the final velocity. The ball initially moves to the left with a velocity of 8.4 m/s, and after bouncing off, it moves to the right with a velocity of 6.8 m/s. So, the change in velocity is 6.8 m/s - (-8.4 m/s) = 15.2 m/s.
Next, we can use the formula for force, which states that force is equal to mass multiplied by acceleration. In this case, the mass of the ball is given as 3.3 kg, and we have calculated the acceleration to be 15.2 m/s. So, the force exerted on the ball by the wall is:
Force = mass x acceleration
Force = 3.3 kg x 15.2 m/s
Force = 50.16 N
Therefore, the constant force exerted on the ball by the wall is 50.16 N.