A 1 kg steel ball strikes a wall with a speed of 9.99 m/s at an angle of 55.1 ◦ with the normal to the wall. It bounces off with the same speed and angle.
If the ball is in contact with the wall for 0.292 s, what is the magnitude of the average force exerted on the ball by the wall? Answer in units of N.
see answer on other post.
To find the magnitude of the average force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of momentum.
First, we need to find the initial and final velocities of the ball after it bounces off the wall.
Given:
Mass of the ball, m = 1 kg
Initial speed of the ball, v1 = 9.99 m/s
Angle of incidence, θ = 55.1 degrees
The initial momentum of the ball can be calculated using the formula:
p1 = m * v1
Next, we use the principle of conservation of linear momentum. In an elastic collision, like in this case, both linear momentum and kinetic energy are conserved.
Since the ball bounces off the wall with the same speed and angle, the final velocity of the ball, v2, can be calculated using the formula:
v2 = 2v1 * cos(θ) / (1 + cos(θ))
Now, we can calculate the final momentum of the ball using the formula:
p2 = m * v2
Using the contact time, Δt = 0.292 s, we can calculate the change in momentum:
Δp = p2 - p1
Finally, we can calculate the magnitude of the average force exerted on the ball by the wall using the formula:
F = Δp / Δt
Plugging in the values and solving the equations, we can find the answer in units of N.