a 200g tennis ball approaches a racket at 20m/s is in contact with the racket for 0.008s and the ball rebounds at 4m/s. find the average force that the racket exerted on the ball
F = Δp/Δt = [m•v2 –(- m•v1)]/ Δt =
= m• (v2+1v1)/ Δt =
0.2(20+4)/0.008 =600 N
To find the average force exerted by the racket on the tennis ball, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the acceleration experienced by the tennis ball during the contact with the racket can be determined using the equation:
a = (v_f - v_i) / t
Where:
- v_f is the final velocity of the ball after the collision (4 m/s)
- v_i is the initial velocity of the ball before the collision (20 m/s)
- t is the duration of the collision (0.008 s)
Substituting the given values:
a = (4 m/s - 20 m/s) / 0.008 s
a = -16 m/s / 0.008 s
a = -2000 m/s²
Note: The negative sign indicates that the direction of acceleration is opposite to the initial velocity.
Now that we have the acceleration, we can calculate the average force applied by the racket:
F = m * a
Substituting the given mass (m = 200 g = 0.2 kg) and the acceleration (a = -2000 m/s²):
F = 0.2 kg * -2000 m/s²
F = -400 N
The average force exerted on the tennis ball by the racket is 400 N. Since force is a vector quantity, the direction of the force is opposite to the initial velocity of the ball.