Question

A ball is falling freely from a tower of height 5m.How much momentum does it transfer to the floor if it's mass is 500g?(a=10m/ssquare). Also chalate kinetic energy transferred by the object

Answer 1

ANSWER

Given,

Height of tower = 5m

Mass of body = 500g =0.5kg

Initial velocity, u = 0m/s

Acceleration, a = 10m/$s^2$

Now

to find the final velocity of body we will use the 3rd law of Kinematics =>

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.

Hence the body will have an kinetic energy of 25J which will be transfered.

Answer 2

ANSWER:-------------

{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:--

T!—!ANKS!!!