A tennis ball of mass 57. g bounces off a wall. Right before hitting the wall, the ball is moving to the right and up with vx=+40. m/s and vy=+10. m/s. Right after bouncing off the wall, the ball is moving to the left and up with vx=-35. m/s and vy=+10. m/s.
If the ball is in contact with the wall for 4.5 ms, what is the average force that the wall makes on the ball?
(I drew the diagrams and tried to use Pythagorean theorem but did not work for some reason.)
thank you.
change of velocity= Vf-vi=-35x+10y-(40x+10y)= -75m/s in x direction
force*time=changemomentum= mass*(-75m/s)
solve for force.
Thanks Bobpurley, I didn't know it was that simple.
To find the average force that the wall exerts on the ball, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum. In this case, since the ball is bouncing off the wall, we can assume that the time of contact is very short, and the change in momentum is equal to the impulse exerted by the wall.
Impulse (J) is equal to the change in momentum (Δp), which can be calculated using the equation:
Δp = m * (vf - vi)
Where:
m is the mass of the ball (in kilograms) = 0.057 kg (converting from grams)
vf is the final velocity of the ball after the collision with the wall
vi is the initial velocity of the ball before the collision with the wall
From the given information, we can see that the x-component of the velocity is changing direction, while the y-component remains the same. Therefore, we can calculate the change in momentum separately for each component.
For the x-component:
Δp_x = m * (vf_x - vi_x)
where:
m = 0.057 kg
vf_x = -35 m/s (final velocity in x-direction)
vi_x = +40 m/s (initial velocity in x-direction)
Δp_x = 0.057 kg * (-35 m/s - 40 m/s)
Simplifying:
Δp_x = -0.057 kg * (-75 m/s)
For the y-component:
Δp_y = m * (vf_y - vi_y)
where:
m = 0.057 kg
vf_y = +10 m/s (final velocity in y-direction)
vi_y = +10 m/s (initial velocity in y-direction)
Δp_y = 0.057 kg * (10 m/s - 10 m/s)
Simplifying:
Δp_y = 0 kg * (0 m/s)
Since there is no change in momentum in the y-direction, Δp_y equals zero.
Now, we can calculate the total change in momentum:
Δp = √(Δp_x² + Δp_y²)
Substituting the values we calculated:
Δp = √((-0.057 kg * -75 m/s)² + (0 kg * 0 m/s)²)
Simplifying:
Δp = √(4.2675 kg*m²/s² + 0 kg*m²/s²)
Now that we have the total change in momentum, we can calculate the average force using the duration of contact:
Average Force (F_avg) = Δp / Δt
where:
Δp = total change in momentum
Δt = duration of contact = 4.5 ms = 4.5 * 10^-3 seconds
F_avg = (Δp) / (Δt)
Finally, substituting the values and calculating:
F_avg = ( √(4.2675 kg*m²/s² + 0 kg*m²/s²) ) / (4.5 * 10^-3 s)
Please note that the calculations provided above assume an elastic collision between the tennis ball and the wall, where no energy is lost during the collision.