An ace footballer converted a penalty shot by kicking the football. Class 9

An ace footballer converted a penalty shot by kicking the football. Class 9

Question 1.

An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h^-1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball. Class 9

Answer:

Converting speed = 108 km h^-1

= 108 × [latex]\frac{1000}{3600}[/latex]

= 30 m s^-1

Initial velocity of ball u = 0

Final velocity v = 30 m s

Using F = [latex]\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}[/latex]

800 = [latex]\frac{0.4 \times(30-0)}{t}[/latex]

800 t = 12 t

= [latex]\frac{12}{800}[/latex]

= 0.015 s

Question 2.

An object of mass 2 kg moving with a constant velocity of 10 m s^-1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest? Class 9

Answer:

Total opposing force = 7 + 3 = 10 N

Using Newton's second law:

a = [latex]\frac{-\mathrm{F}}{\mathrm{~m}}[/latex]

= [latex]\frac{-10}{2}[/latex]

= -5 m s^-2 (deceleration)

Initial velocity u = 10 m s^-1

final velocity v = 0

Using v² = u² + 2as

0 = (10)² + 2 × (-5) × s

= 100 - 10s s

= 10 m

The object travels 10 m before coming to rest.