Question

How much momentum will dum bell of mass 10kg transfer to the floor If it fall from a height of 80cm ?take downward acceleration as 10m/s^2

Answer 1

Answer:

40kg m/s

Explanation:

Given

Mass of the dúmb bell,m = 10kg

Distance covered by the dúmb-bell,s= 80 cm = 0.8m       [1cm =1/100m]

Acceleration in the downward direction,a = 10 m/s^2

Initial velocity of the dúmb-bell,u = 0

To Find :

Momentum,p = ?

Solution :

We know,

Momentum (p) = Mass (m) × velocity (v)

We don’t know it’s  velocity,so let’s find out it’s velocity first.

According to third equation of motion

$\sf{}v^2=u^2+2as$

$\sf{}\Rightarrow v^2=0+2 \times 10ms^{-2}\times 0.8m$

$\sf{}\Rightarrow v^2=16m/s$

$\sf{}\Rightarrow v^2=16m^2s^{-2}$

$\sf{}\Rightarrow v=\sqrt{16m^2s^{-2}}$

$\sf{}\therefore v=4m/s$

Now,we know both velocity and mass

Therefore moentum of the dùmb-bell:

$\sf{}\Rightarrow 10kg\times 4m/s$

$\sf{}\therefore 40kg\ m/s$

Answer 2

$\sf{\pink{\underline{\underline{\purple{GIVEN:-}}}}}$

• Mass (m) = 10kg

• Height (h) = 80cm = 0.8m

• Acceleration (a) = 10m/s²

$\sf{\pink{\underline{\underline{\purple{TO\: FIND:-}}}}}$

• Momentum = ??

$\sf{\pink{\underline{\underline{\purple{SOLUTION:-}}}}}$

We have know that,

• initial velocity (u) = 0m/s²

$\orange\star\:\bf{\gray{\boxed{\boxed{\red{v^2\:=\:u^2\:+\:2as\:}}}}}$

Where,

• v = final velocity

• u = 0m/s

• a = 10m/s²

• s = 0.8m

$\rm{\implies\:v^2\:=\:0^2\:+\:2\times{10}\times{0.8}\:}$

$\rm{\implies\:v^2\:=\:0\:+\:20\times{0.8}\:}$

$\rm{\implies\:v\:=\:\sqrt{16}\:}$

$\rm\pink{\implies\:v\:=\:4\:m/s\:}$

☃️ Now, Momentum transferred to ground,

$\green\star\:\bf{\gray{\boxed{\boxed{\red{p\:=\:mv\:}}}}}$

Where,

• p = momentum

• m = 10kg

• v = 4m/s

$\rm{\implies\:p\:=\:10\times{4}\:}$

$\rm\pink{\implies\:p\:=\:40\:kg.m.s^{-1}\:}$

$\rm\green{\therefore}$ Momentum transferred to the ground is “40 kg.m/s ”