Question

if a dummb-bell of mass 10 kg falls from a height of 1.8 m, how much momentum will get transferred to the floor? You may take g = 10 m/s^2

Answer 1

Answer:-

Known Terms:-

m = Mass

a = Acceleration

v = Final Velocity

u = Initial Velocity

s = Distance

Given:-

Mass of dummb-bell, m = 10 kg

Acceleration by Dummb-bell, g = 10 m/s²

Initial Velocity of dummb-bell, u = 0

Final Velocity of dummb-bell, v = ???

Distance Attain by the dummb-bell, s = 1.8 m

To Calculate:-

Final Velocity

Momentum Transferred

Formula to be used:-

3rd equation of motion that is v² - u² = 2 gs   (∵ a = g)

Momentum Formula that is p = mv

Solution:-

Final Velocity

Putting all the values we get,

⇒ v² - u² = 2 gs

⇒ v² - (0)² = 2 × 10 × 1.8

⇒ v² = 36

⇒ v = $\sqrt{36}$

⇒ v = 6 m/s

Momentum Transferred

Putting all the values, we get

⇒ Momentum = Mass × Final Velocity

⇒ Momentum = 10 × 6

⇒ Momentum = 60 kg m/s.

Hence, Momentum Transferred to the floor is 60 kg m/s.

Answer 2

Answer: 60 kg m /s

Explanation:

Given,

Mass of the bell =  10 kg

height attained by the bell = 1.8 m

As we know ,

v = final velocity

u = initial velocity

s = distance covered

p = momentum

from the equation of motion,

$v^{2}-u^{2}=2gs$

here,

u = 0

s = 1.8 m

g = 10$m/s^{2}$

⇒ v² - u² = 2 gs

⇒ v² - (0)² = 2 × 10 × 1.8

⇒ v² = 36

⇒ v = $\sqrt{36}$

⇒ v = 6 $m/s$

∴ final velocity = 6$m/s$

Momentum transferred =  mass (m) × velocity (v)

                             

                                         = 10 × 6

                                         = 60 $kgm/s$