A foot ball of mass 0.5 kg moving with a velocity
of 10 m s-1 hits a pole and, the ball turns back and
moves with a velocity of 20 m s-1. Find the impulse
and the force exerted on the ball if the force acts for
0.02 s.
Answers
Given,
m = 0.5 kg
t = 0.02 s
u = 10 m s^(-1)
After collision with the pole, the ball turns back so the velocity will be in opposite direction. Hence,
v = - 20 m s^(-1)
We know that impulse of a body is the change in momentum of that body. So,
Impulse = Final momentum - Initial momentum, i.e.,
I = m (v - u)
I = 0.5 (- 20 - 10)
I = 0.5 × (- 30)
I = - 15 kg m s^(-1)
And the force exerted on the ball is the rate of change of momentum, or the impulse imparted per unit time, i.e.,
F = I / t
F = - 15 / 0.02
F = - 750 N
Answer
Momentum of ball is changed by 7.5 kg m/s.
A force of 375 N is exerted on the ball.
Explanation
Given:-
Mass of football, m = 0.5 kg
Initial velocity of football, u = 10 m/s
Ball hits a pole and bounces back with certain velocity
Final velocity of football, v = -5 m/s [Negative because ball bounces back]
time for which force acts, t = 0.02 s
To find:-
Change in momentum of the ball, Δp =?
Force exerted on the ball, F =?
Formulae required:-
Formula for Change in momentum
Δp = (Final p) - (Initial p) = m v - m u
Formula for force
F = Δp / t
Where, p is momentum, m is mass, u is initial velocity, v is final velocity, t is time.
Solution:-
Calculating the change in momentum of ball
→ Δp = (Final p) - (Initial p) = m v - m u
→ Δp = m v - m u
→ Δp = m ( v - u )
→ Δ p = 0.5 ( -5 -10 ) = -7.5 kg m/s
Therefore,
The magnitude of change in momentum of the ball is 7.5 kg m/s.
Now, Calculating force exerted on the ball
→ F = Δp / t
→ F = (-7.5) / 0.02
→ F = -375 N
Therefore,
A force of magnitude 375 N is exerted on the ball by the wall.