A 4.02 kg steel ball strikes a massive wall at 11.8 m/s at an angle of 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle, as shown in the figure below. If the ball is in contact with the wall for 0.174 s, what is the magnitude of the average force exerted by the wall on the ball?
x-axis is along the ball motion and nornal to the wall , y-axis is parallel to the wall
v=v₀ =11.8 m/s ,
v(x)=v•sin60°=11.8•0.866=10.22 m/s
linear momentum change is Δp
Δp(y) = m•Δv(y) =m[v-v₀]= 0.
Δp(x)=m•Δv(x) =m[v-(-v₀)] = m•2•v₀ = F•Δt
F= 2•m•v₀/Δt = 2•4.02•10.22/0.174=472.18 N
To find the magnitude of the average force exerted by the wall on the ball, we can use the principle of impulse-momentum. Impulse is defined as the change in momentum of an object, and it is equal to the average force multiplied by the time the force is applied.
First, let's find the initial momentum of the ball before it strikes the wall. The momentum of an object is equal to its mass multiplied by its velocity.
The mass of the ball, m = 4.02 kg.
The initial velocity of the ball, v = 11.8 m/s.
Using the given angle of 60.0°, we can calculate the vertical and horizontal components of the initial velocity.
The vertical component of velocity, v_y = v * sin(angle)
v_y = 11.8 * sin(60.0°)
v_y ≈ 11.8 * 0.866 ≈ 10.206 m/s
The horizontal component of velocity, v_x = v * cos(angle)
v_x = 11.8 * cos(60.0°)
v_x ≈ 11.8 * 0.5 ≈ 5.900 m/s
Since the ball bounces off the wall with the same speed and angle, we can conclude that the vertical component of its velocity remains the same, but the horizontal component will be negative.
The final vertical component of velocity, v'_y = 10.206 m/s
The final horizontal component of velocity, v'_x = -5.900 m/s
Now, let's find the final momentum of the ball after it bounces off the wall.
The final velocity of the ball can be found by combining the horizontal and vertical components.
The final velocity, v' = √(v'_x^2 + v'_y^2)
v' = √((-5.900)^2 + (10.206)^2)
v' ≈ √(34.81 + 104.27)
v' ≈ √139.08 ≈ 11.79 m/s
The change in momentum, Δp = m * (v' - v)
Δp = 4.02 * (11.79 - 11.8)
Δp ≈ 4.02 * (-0.01) ≈ -0.0402 kg*m/s
Since the impulse is equal to the change in momentum, we can calculate the average force using the following equation:
Impulse = Average Force * Time
Impulse = Δp = -0.0402 kg*m/s (negative sign indicates the change in direction)
Time, t = 0.174 s (given)
Therefore, we can find the average force:
Average Force = Δp / t
Average Force = -0.0402 kg*m/s / 0.174 s
Calculating the average force:
Average Force ≈ -0.0402 kg*m/s / 0.174 s ≈ -0.2312 N
The magnitude of the average force exerted by the wall on the ball is approximately 0.2312 N.