How Does Kicking a Soccer Ball Affect Its Momentum and Force?

[songoku]

Homework Statement

A ball with mass 100 gr is kicked with angle -30^o to the ground and has speed 20 ms^-1. Then, the ball is kicked and reversed with angle 30^o 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

Homework Equations

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

The Attempt at a Solution

I'm not very sure about the question at part "the ball is kicked and reversed with angle 30^o". 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

 

[Hootenanny]

I'm not very sure about the question at part "the ball is kicked and reversed with angle 30^o".

I would read this statement as "the ball is initially traveling 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.

 

[songoku]

Hi Hootenanny

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

Thanks

 

[Hootenanny]

Hi Hootenanny

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

Thanks

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.

 

[songoku]

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:
*V_x = 20 cos 30^o to the right
*V_y = 20 sin 30^o downward

b. When the ball kicked:
*V_x = 25 cos 30^o to the right
*V_y = 25 sin 30^o upward

P_x = 25 cos 30^o-20 cos 30^o=5 cos 30^o
P_y = 25 sin 30^o - (-20 sin 30^o) = 45 sin 30^o

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

 

[Hootenanny]

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.

 

[songoku]

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

 

[Hootenanny]

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

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.

 

[songoku]

Hi Hootenanny

Ok now I get it.

Thanks a lot for your help :)