Question

how much Momentum will a dump bell of mass 10kg transfer to the floor if it falls from a height of 80 CM take its downward acceleration to be 10 metre per second square​

Answer 1

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p=?

mass is 10 kg

h is 80 cm =0.8m

acceleration is 10 m/

u =o since downward acceleration

v=?

by using third law of motion

v*2=u*2+2as

v*2= 0+16

v=4

there--- momentum's

is =m×v

is 40kgm/s

Answer 2

$\star{\underline{\underline{\sf \red{ Solution:-} }}}$

Given:

• Mass Of The bell (M) = 10kg
• Distance covered (s) = 80cm(0.8m)
• Initial velocity (u) = 0
• Acceleration (a) = $\sf{10ms^{-2}}$

$\\ \\ \star{\underline{\underline{\sf \red{ To\:Find:-} }}}$

• Terminal-velocity (v) =?

$\\ \star{\underline{\underline{\sf \red{ Solution:-} }}}$

Momentum of the bell when it hits the ground = mv

According to the third Motion Equation,

$\bullet{\boxed{\sf{ V^2 = u^2 + 2as}}}$

After substituting the values,

$\longrightarrow\sf{ V^2 = 0 + 2 \times (10ms^{-2}) (0.8m) } \\ \\ \longrightarrow\sf{ v^2 - 0 = 2 \times (10ms^{-2}) (0.8m) } \\ \\ \longrightarrow\sf{v^2 = 16\:m^2s^{-2}}$

$\longrightarrow{\sf{ v = \sqrt{ 16} }} \\ \\ \longrightarrow\sf{ 4 \: ms^{-1} }$

Now,

The momentum transferred by the bell to the floor;

$\implies\sf{ (10kg) \times (4 m/s)} \\ \\ \implies\sf{ 40 kg\:ms^{-1}}$