Question

A body of mass 2 kg is dropped from a height of 20 m. It accelerates uniformly at the rate of 10 m/s2 until it hits the ground. Calculate the initial momentum and the momentum just before it strikes the ground.

Answer 1

s= ut + 1/2 at × t

and we have m= 2 kg

u = 0

h = 20m

a = 10m/s2

and initial momentum = ?

and momentum just before it strikes the ground = ?

now ,

h = ut + 1/2 gt2

now ,20= (o × t + 1/2 × 10 × t2)

20 = 5t

t= 4s

now , v = u + at

so ,v = (o + 10 × 4)

v = 40 m/ s

now , initial momentum = m × v

= 2 × 40

= 80 m/s

and momentum = m × a

= 2 × 10

= 20 kg.ms

Answer 2

Answer:

20 kgms

Explanation:

s= ut + 1/2 at × t  

m= 2 kg     u = 0     h = 20m     a = 10m/s2   h = ut + 1/2 gt2

now ,20= (o × t + 1/2 × 10 × t2)

20 = 5t

t= 4s

now , v = u + at

v = (o + 10 × 4)

v = 40 m/s

initial momentum = m × v

= 2 × 40

= 80 m/s

momentum = m × a

= 2 × 10

= 20 kgms