physics homework problem on impulse

Question

A punter drops a ball from rest vertically 1 meter down
onto his foot. The ball leaves the foot with a speed of 18 m/s
at an angle 55º above the horizontal. What is the impulse
delivered by the foot (magnitude and direction)? The mass of the ball is .41 kg.

Answer 1

Hi Anakin.

 

This is an impulse-momentum problem. An "impulse" is defined as a "change in momentum." That means we have to find the momentum of the ball before it was kicked, and the momentum after it was kicked, and then calculate their difference.

 

Momentum = mass*velocity, so we have some velocities to figure out. Let's declare up and to the right as positive.

 

The ball fell a meter at 9.8 m/s^2.

 

d = 1/2*g*t^2

 

1 = 4.9*t^2

 

t^2 = 1/4.9

 

t = 0.45 sec

 

9.8 m/sec^2 * 0.45 sec = 4.43 m/sec downward = -4.43 m/sec j

 

The mass of the ball is 0.41 kg.

 

Initial Momentum of the ball (Pi) before kick is:

 

Pi = 0.41kg*(-4.43 m/sec j)

 

= -1.82 N*sec j.

 

The problem then gets a little tricky when we think about the kick. Our initial momentum only has a vertical component, but when the ball is kicked, it has both a vertical component and a horizontal component as given by the 55º angle. Since momentum is a vector quantity, we can't discount the horizontal component.

 

It's launched with a velocity of 18 m/sec at an angle of 55º. That means that:

 

Vx = 18*cos(55)

Vy = 18*sin(55)

 

and thus the final momentum (Pf) is:

 

Pf_x = 0.41*18*cos(55) = 4.23 N*sec i

Pf_y = 0.41*18*sin(55) = 6.05 N*sec j

 

The change in momentum is Pf - Pi:

 

(4.23 i + 6.05 j) - (-1.82 j) = 4.23 i + 7.87 j

 

Now do a vector sum:

 

√(4.23² + 7.87²) = 8.93 N*sec

 

To find the angle, recognize that tan = y/x, so arctan(y/x) = theta

 

arctan(7.87/4.23) = ~62 degrees

 

8.93 N*sec at an angle of 62 degrees.

 

This is your impulse.

 

Hope this helps.

Answer 2

Anakin,

We typically think of impulse as being the force multiplied by time, but no time is given. In this case, we look at impulse as being the change in momentum, or I = m*Δv. Since Δv = v_f - v₀, we need to evaluate v_f and v₀ in vector form so that they can be combined.

 

for v₀, we use the gravity equation v² = 2a(Δh)

 

v₀ = √(19.6*1) = 4.43 m/s, negative because it is falling.

 

v₀ = 0i - 4.43j

 

v_f = 18cos(55)i + 18sin(55)j = 10.32i + 14.74j

 

I = .41[(10.32i + 14.74j) - (0i - 4.43j)]

I = .41[10.32i + 19.17j]

I = (4.23i + 7.86j) N*s     Now apply Pythagorean theorem for magnitude and arc-tangent for direction:

 

√(4.23² + 7.86²) = 8.93N*s

 

arctan(7.86/4.23) = 61.7º above the horizontal.