A soccer ball of mass 450 grams is initially moving horizontally east at 12 m/s. You kick it so that it moves horizontally north at 13 m/s. The collision lasts 0.04 s. Find the magnitude and direction of the average force you apply to the ball.
|F| =
Angle from East =
not right
Divide the momentum change vector by 0.04 s.
The magnitude of the momentum change vector is
sqrt[(12)^2 + (13^2] = 17.69 N*s
Next, do the division.
The direction is the direction of the momentum change.
arctan13/12 = 47.3 degrees from east
To find the magnitude of the average force applied to the ball, we can use the impulse-momentum theorem. The impulse experienced by an object is given by the change in its momentum:
Impulse = Δp = mΔv
where m is the mass of the ball and Δv is the change in velocity.
In this case, the mass of the ball is 0.45 kg (450 grams) and the change in velocity in the north direction is 13 m/s - 12 m/s = 1 m/s.
Impulse = 0.45 kg * 1 m/s = 0.45 kg⋅m/s
Since impulse is equal to the average force multiplied by the time interval, we have:
Impulse = FΔt
where F is the average force and Δt is the time interval.
Plugging in the values, we have:
0.45 kg⋅m/s = F * 0.04 s
Solving for F:
F = (0.45 kg⋅m/s) / (0.04 s) = 11.25 N
Therefore, the magnitude of the average force applied to the ball is 11.25 Newtons.
To find the angle from East, we can use trigonometry. Since the ball initially moves horizontally east and ends up moving horizontally north, the angle between the initial and final velocities is 90 degrees.
Thus, the angle from East is 90 degrees.
To find the magnitude and direction of the average force applied to the ball, we can use Newton's second law of motion, which states that force is equal to the change in momentum divided by the change in time.
The momentum of an object is defined as the product of its mass and velocity. In this case, the momentum before the collision is given by the product of the mass (450 grams) and the initial velocity (12 m/s) in the eastward direction, and the momentum after the collision is given by the product of the mass (450 grams) and the final velocity (13 m/s) in the northward direction.
Before the collision:
Initial momentum = mass × initial velocity = 0.45 kg × 12 m/s
After the collision:
Final momentum = mass × final velocity = 0.45 kg × 13 m/s
The change in momentum is the difference between the final momentum and initial momentum:
Change in momentum = Final momentum - Initial momentum
To calculate the magnitude of the average force, we divide the change in momentum by the time of collision:
|F| = (Change in momentum) / (collision time)
Now let's calculate the magnitude of the average force:
Change in momentum = (0.45 kg × 13 m/s) - (0.45 kg × 12 m/s)
= 0.45 kg × (13 m/s - 12 m/s)
Collision time = 0.04 s
|F| = (0.45 kg × (13 m/s - 12 m/s)) / 0.04 s
Now let's calculate the magnitude of the average force:
|F| = (0.45 kg × 1 m/s) / 0.04 s
Finally, we can calculate the magnitude of the average force applied to the ball:
|F| = (0.45 kg × 1 m/s) / 0.04 s
|F| = 11.25 N
So, the magnitude of the average force applied to the ball is 11.25 N.
To find the direction of the average force, we can use trigonometry. The angle from the east can be calculated using the relationship between the horizontal and vertical components of the force:
tan(angle from east) = (Final vertical velocity) / (Initial horizontal velocity)
Now let's calculate the angle from the east:
angle from east = arctan((Final vertical velocity) / (Initial horizontal velocity))
= arctan(13 m/s / 12 m/s)
Finally, we can calculate the angle from the east:
angle from east = arctan(13 / 12)
angle from east ≈ 47.41°
So, the magnitude of the average force applied to the ball is 11.25 N and the angle from the east is approximately 47.41°.