Question

...A dum-bell of 5kg mass falls from a height of 0.4m ?what is the momentum transferred by the the dum while hitting the ground?​

Answer 1

A dum-bell of mass 5kg mass falls from a height of 0.4m. What is the momentum transferred by the dum-bell while hitting the ground?​

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Mass (m) = 5 Kg.
• Height (s) = 0.4 m.
• Acceleration due to gravity (g) = 9.8ms⁻².
• Initial velocity (u) = 0ms⁻¹.

$\LARGE{\bf{\underline{\underline{TO:FIND:-}}}}$

• Momentum (ρ)

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Recall the three equations of motion.

$\boxed{\substack{\displaystyle \sf v=u+at \\\\ \displaystyle \sf s=ut+\dfrac{1}{2}gt^2 \\\\ \displaystyle \sf 2as=v^2-u^2}}$

It is apt to use the third formula because we are given u, s and a and we need to find v.

$\to \sf 2as=v^2-u^2$

Substituting the values:

$\sf \to 2 \times 9.8 \times 0.4= v^2-0^2$

$\sf \to v=\sqrt{7.84}$

$\boxed{\sf{\green{v=2.8ms^{-1}}}}$

We know that:

$\sf{\red{Momentum(\rho) = mass (m) \times velocity(v)}}$

Substituting the values, we get:

$\to \rho=5 \times 2.8$

$\boxed{\sf{\green{\rho=14 k g m / s}}}$

MOMENTUM - 14 Kg · m/s