# Momentum of soccer ball

1. Aug 28, 2009

### songoku

1. The problem statement, all variables and given/known data
A ball with mass 100 gr is kicked with angle -30o to the ground and has speed 20 ms-1. Then, the ball is kicked and reversed with angle 30o and speed 25 ms-1. The player touches the ball for 0.1 s. Find:
(i) Change in momentum
(ii) Force by the player
(iii) Work by the player
(iv) Power exerted on the ball

2. Relevant equations
p = momentum = mv = F.t
P = power = W/t
Work = change in kinetic energy

3. The attempt at a solution
I'm not very sure about the question at part "the ball is kicked and reversed with angle 30o". I tried to solve this problem with assumption that the ball is kicked back in a straight line.

(i) change in momentum = mv-mu = 0.1 (25 - (-20) ) = 4.5 Ns

(ii) $$F=\frac{p}{t}=\frac{4.5}{0.1}=45N$$

(iii) W = 1/2 mv2 - 1/2 mu2 = 11.25 J

(iv) P = W/t = 11.25 / 0.1 = 112.5 Watt

Do I get it right ?

Thanks

2. Aug 28, 2009

### Hootenanny

Staff Emeritus
Re: Momentum

I would read this statement as "the ball is initially travelling with speed 20m/s at an angle of -30 degrees. The ball is then kicked and travels at an angle of 30 degrees and speed 25m/s".

In this case, the motion is not rectilinear.

3. Aug 28, 2009

### songoku

Re: Momentum

Hi Hootenanny

So you mean that the motion of the ball is like "V" letter?

Thanks

4. Aug 28, 2009

### Hootenanny

Staff Emeritus
Re: Momentum

Technically, a sidewards "V" since the angles are given with respect to the ground, but yes I would say that the trajectory of the ball is V-shaped.

5. Aug 28, 2009

### songoku

Re: Momentum

Hi Hootenanny

Ok so I've tried new solution.

(i) Because the trajectory of the ball is V-shaped, I broke the velocity onto two components.

a. When the ball comes:
*Vx = 20 cos 30o to the right
*Vy = 20 sin 30o downward

b. When the ball kicked:
*Vx = 25 cos 30o to the right
*Vy = 25 sin 30o upward

Px = 25 cos 30o-20 cos 30o=5 cos 30o
Py = 25 sin 30o - (-20 sin 30o) = 45 sin 30o

Change in momentum = $$\sqrt{(P_x)^2+(P_y)^2} = 5\sqrt{21}\;Ns$$

(ii) $$F=\frac{p}{t}=\frac{5\sqrt{21}}{0.1}=50\sqrt{21} \;N$$

(iii) W = 1/2 mv2 - 1/2 mu2 = 11.25 J

(iv) P = W/t = 11.25 / 0.1 = 112.5 Watt

Do I get it right?

Thanks

Last edited: Aug 28, 2009
6. Aug 28, 2009

### Hootenanny

Staff Emeritus
Re: Momentum

I haven't checked your arithmetic, but your method looks fine two.

One, somewhat minor point: since we are dealing with planar motion, both the change in momentum and the force should be vector quantities.

7. Aug 28, 2009

### songoku

Re: Momentum

Hi Hootenanny

I've re-checked my answer and I have doubt at (ii). The question is asking about the force by the player, so should we only consider the momentum when the ball kicked with speed 25 ms-1 because the player doesn't give any forces when the ball comes?

Thanks

8. Aug 29, 2009

### Hootenanny

Staff Emeritus
Re: Momentum

Newton's second law states that the net external force acting on a body is equal to the rate of change of the momentum. Therefore, you need to consider the change in momentum, not just the out-going momentum.

9. Aug 29, 2009

### songoku

Re: Momentum

Hi Hootenanny

Ok now I get it.

Thanks a lot for your help :)