Impulse Question Plz need help ASAP

Question

A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 27 m/s at an angle 60° above the horizontal. What is the impulse delivered by the foot? (Use 0.41 kg for the mass of the ball. Assume the punter faces the +x-direction and that the +y-direction is upward. Enter the impulse in kg · m/s and the direction in degrees counterclockwise from the +x-axis.)

Answer 1

This is about conservation of momentum. Keep in mind that momentum is a vector quantity. The final momentum of the ball is the vector sum of the initial momentum and the impulse imparted by the kicker.

To get the initial momentum, we need the velocity of the ball before it is kicked. Use d = 1/2 * a * t^2 to find the time to drop 1 meter. The velocity is then a*t. The momentum is then the mass times the velocity. (Direction is -90 degrees, using the coordinate system given)

The easiest way to add the vectors is by first finding the x and y components. Again using the specified coordinate system, the initial momentum has only a y component, and it is negative.

The final velocity of 27 m/sec resolves into an x component of 27cos(60), and a y component of 27sin(60)

Combine the vectors:

The only x component is 27cos(60), so this is the x component of the kick. The y component is SUM of the initial downward momentum and 27cos(60).

Once you have the final momentum components, determine the magnitude and direction using the trig rules

Answer 2

Hi José,

The question asks for the impulse that the punter's foot exerts on the ball. At its core, Impulse = change in momentum and since this is a 2D problem we can apply the equation for momentum to the x and y directions.

I_x = m_fv_x,f - m_iv_x,i

I_y = m_fv_y,f - m_iv_y,i

For your problem:

We can assume that the mass of the ball remains constant, so

m_i = m_f = 0.41 kg.

We know that the ball was initially traveling straight down, so

v_x,i = 0 m/s.

We know the final velocity of the ball is 27 m/s at 60 deg which allows us to solve for

v_x,f = 27*cos(60) = 13.5 m/s and

v_y,f = 27*sin(60) = 23.3827 m/s.

Plug all this into I_x and I_y

I_x = 0.41*13.5 - 0.41*0 = 5.535 kg*m/s

I_y = 0.41*23.3827 - 0.41v_y,i

At this point, you should notice that we are still missing v_y,i and require this unknown to find the impulse. Since we were told the ball was dropped 1 m from rest we can calculate v_y,i using a basic kinematic equation (assuming air resistance is not an issue).

v² = v₀² + 2a(Δx)

v_y,i² = 0² - 2*9.81*(-1)

v_y,i = 4.42945 m/s downwards

Now we can finish solving for I,

I_x = 5.535 kg*m/s

I_y = 0.41*23.3827 - 0.41*(-4.42945) = 11.403 kg*m/s

Using vector math, we know that the magnitude of I can be solved using,

|| I || = sqrt(I_x² + I_y²)

|| I || = sqrt(5.535² + 11.403²) = 12.6753 kg*m/s.

Also, the angle of I can be solved using,

θ_I = arctan(I_y / I_x)

θ_I = arctan(11.403 / 5.535) = 64.108 deg

Final answer: I = 12.68 kg*m/s at an angle of 64.12 deg.

I hope that helps! If you have any questions let me know and I will be glad to lend a hand.

Bryce