Question

a ball is falling freely from a tower of height 5m. how much momentumdoes it transfer to the floor if its mass is 500g? (take g = 10 meter/sec square)​

Answer 1

Answer: 25 joule

Explanation:

V^2 - u^2 = 2a×s

=> V^2 - 0 = 2(10)×(5)

=>V^2 = 100

=> V = 10m/s.

Hence the final velocity of body is

10m/s.

Now we know that =>

Momentum, p = mass × velocity

=>p = m × v

=>p = 0.5 × 10

=>p = 5kg m/s

Hence the momentum of body is

5kgm/s.

Now we know that =>

Kinetic energy = 1/2 × m × v^2

=>K. E = 1/2 × 0.5 × 100

=> K. E = 25 joules.

Answer 2

$\huge{\underbrace{\overbrace{\mathfrak{\pink{Answer:}}}}}$

given:-

• {Height of tower = 5m}

• { body = 500g =0.5}

• {velocity, u = 0m/s}

• {Acceleration, a = 10m/s2s^2s2}

velocity of body we will use the 3rd law

=>V^2 - u^2 = 2a×s

• => V^2 - 0 = 2(10)×(5)

• =>V^2 = 100

• => V = 10m/s.

velocity of body is10m/s.

Now we know that =>

• Momentum, p

• = mass × velocity
• =>p = m × v
• =>p = 0.5 × 10
• =>p = 5kg m/s

momentum of body is5kgm/s

we know that =>Kinetic energy =

• 1/2 × m × v^2=>K. E = 1/2 × 0.5 × 100=> 

• K. E = 25 joules.

hence proved:)

hope it helps:--