A 2 kg steel ball strikes a wall with a speed of 8.15 m/s at an angle of 54.4◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. If the ball is in contact with the wall for 0.243 s, what is the magnitude of the average force exerted on the ball by the wall?
Answer in units of N
Divide the momentum change magnitude by 0.243 seconds.
I would need to see the figure to know what the momentum change is. It could bounce off in the same direction that it came (backwards), or in a 54.4 degree direction on the other side of the normal.
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 force (F) is equal to the rate of change of momentum. In mathematical terms, this can be written as:
F = (mv - mu) / t,
where F is the force, m is the mass of the ball, v is the final velocity of the ball, u is the initial velocity of the ball, and t is the time interval.
First, we need to find the initial and final velocities of the ball.
Given:
Mass of the ball (m) = 2 kg
Initial velocity of the ball (u) = 8.15 m/s (at an angle of 54.4 degrees)
Time interval (t) = 0.243 s
To find the initial and final velocities, we need to split the initial velocity into horizontal and vertical components. We can use trigonometry to find these components.
Vertical component:
Initial velocity in the vertical direction (uy) = u * sin(angle)
uy = 8.15 m/s * sin(54.4 degrees)
Horizontal component:
Initial velocity in the horizontal direction (ux) = u * cos(angle)
ux = 8.15 m/s * cos(54.4 degrees)
Since the ball bounces off with the same speed and angle as it hits the wall, the final velocity will be the negation of the initial velocity.
Final velocity in the vertical direction (vy) = -uy
Final velocity in the horizontal direction (vx) = -ux
Now we have the initial and final velocities:
Initial velocity (u) = sqrt(ux^2 + uy^2)
Final velocity (v) = sqrt(vx^2 + vy^2)
Next, we can substitute the values into the equation for force:
F = (mv - mu) / t
F = (m * (sqrt(vx^2 + vy^2)) - m * (sqrt(ux^2 + uy^2))) / t
Now we can calculate the magnitude of the average force by substituting the given values into the formula:
F = (2 kg * (sqrt((-8.15 m/s * cos(54.4 degrees))^2 + (-8.15 m/s * sin(54.4 degrees))^2)) - 2 kg * (sqrt((8.15 m/s * cos(54.4 degrees))^2 + (8.15 m/s * sin(54.4 degrees))^2))) / 0.243 s
After evaluating the above expression, you will get the magnitude of the average force exerted on the ball by the wall in units of Newton (N).