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 -625 N.

Explanation:

We have,

• Mass (m) = 0.5 Kg
• Initial velocity (u) = 10 m/s
• Final velocity (v) = 15 m/s
• Contact time (t) = 0.02 sec

To find,

• Average force (F)

Solution,

Using the formula,

=> Force = Chane in momentum/Time taken

Change in momentum:

>> P = mv

=> P = 0.5 × (-15 - 10)

=> P = 0.5 × 25

∴ Momentum = -12.5 Kg.m/s

Now,

=> Force = -12.5/0.02

∴ Force = -625 N

Answer 2

GIVEN:-

• 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

TO FIND:-

The average force involved (F)

EXPLANATION:-

From newton's 2nd law ,

>f=m(v-u)/t

>F = 0.5(-25)/0.02

> F = -12.5/0.02

> F = -1250/2

> F = -625 N

Negative sign means retarding force .

EXTRA INFORMATION:-

Derivation of F-ma

The rate of change of momentum of a body is directly proportional to applied force on it :-

Rate of change of momentum=∝ΔP/t

ΔP=change in momentum

t=time

F=force

F∝Pf-Pi/t

F∝mv-mu/t

F∝m(v-u/t)

F∝ma                     [(v-u/t)=a]

F=kma

K=1 (experimentally)

F=MA