Question

A football of mass 0.5 kg coming towards player with a velocity of 10 m/s. Player hit straight back the football with velocity 15 m/s. If the contact time is 0.02 second then the average force involved is: - (1) 250 N (2) 1250 N (3) 500 N (4) 625 N​

Answer 1

Answer :-

Average force involved is a retarding force of 625 N [Option.4] .

Explanation :-

We have :-

→ Mass of the football (m) = 0.5 kg

→ Initial velocity (u) = 10 m/s

→ Final velocity (v) = 15 m/s

→ Contact time (t) = 0.02 sec

______________________________

We can understand that initially the football was moving in one direction, and the player hit it back to the opposite direction. So, it's final velocity will be -15 m/s .

Now, let's calculate the average force by using Newton's 2nd law of motion .

F = m(v - u)/t

⇒ F = 0.5(-15 - 10)/0.02

⇒ F = 0.5(-25)/0.02

⇒ F = -12.5/0.02

⇒ F = -1250/2

⇒ F = -625 N

[Here, -ve sign represents retarding force ].

Some Extra Information :-

The quantity of motion contained in a body is called "momentum" of the body .

In this case :-

• Initial momentum of ball = m × u

• Final momentum of ball = m × v

Answer 2

Given:-

A football of mass 0.5 kg coming towards a player with a velocity of 10 m/s. The player hit straight back the football with a velocity of 15 m/s.

To Find:-

The Average force

Solution:-

We know that

${\boxed{\frak{\pink{\underline{\underline{\ast a=\dfrac{v-u}{t}\ast}}}}}}$

Where

a = acceleration

v = final velocity

u = initial velocity

t = time

$\sf :\implies a=\dfrac{-15-10}{0.02}$

$\sf :\implies a=\dfrac{-25}{0.02}$

$\sf :\implies a=\dfrac{-2500}{2}$

$\sf :\implies a=-1250\;ms^{-2}$

Now, We need to find the average force.

${\boxed{\frak{\pink{\underline{\underline{\ast F=ma\ast}}}}}}$

Where

F = force

m = mass

a = acceleration

$\sf :\implies F=0.5\times(-1250)$

$\sf :\implies F=\dfrac{5}{10}\times -1250$

$\sf :\implies F=5\times -125$

$\sf :\implies F=-625\;N$