Question

A d_umbbell of mass 10 kg falls on the floor from a height of 80 cm. Calculate the change in momentum of the d_umb bell and also the force of impact on the floor.

(Take g= 10 m/s²)​

Answer 1

Explanation:

Hola !

we know the formula ,

v²-u²=2as

Here intinal velocity is zero .

such that

v²-0²=2×10×0.80

v²=20×0.80

v²=16

v=4

And momentum is given by p =MV

p=10×4

p=40

Thank you !

#Astro

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\pink{Answer:-}$

⏩ Here, u= 0, g= 10 m/s², h= 80 cm= 0.8 m.

We know that,

v² = u² + 2gh

=> v² = 0 + 2 × 10 × 0.8

=> v² = 16

=> v= 4 m/s

→Change in momentum = mv - mu

= 10 kg × 4 m/s

= 40 kg m/s

Now,

**Time taken by the bell to reach the floor:

v = u + gt

=> t= 4/10 = 0.4 s

Therefore,

→Force on the floor , F = m(v-u)/t

= (10×4)/0.4

= 100 N.

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Hope it helps...:-)

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