Question

# How much momentum will a dumb-bell of mass 10kg transfer to the floor if it falls from a height of 80 cm? take its downward acceleration to be 10 $m{{s}^{-2}}$.

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Hint: In solving this question, first we will use the third equation of motion given by Newton and use it to find the value of the final velocity of the dumb-bell and then use the velocity we get in the calculation of the momentum of the body and that will be the momentum that will be transferred to the floor.

Formula used: $2as={{v}^{2}}-{{u}^{2}}$

Complete step-by-step solution:
We will assume that when the dumb-bell falls to the floor, it does not bounce and remains on the floor i.e. all of its momentum gets transferred to the floor at the same moment. Now we will use the third equation of motion by Newton to find what will be the velocity of the dumb-bell just before it strikes the ground.
$2as={{v}^{2}}-{{u}^{2}}$
Here, a is the acceleration on the body (dumb-bell here) which here will be the acceleration due to gravity. Its magnitude has been given as 10 $m{{s}^{-2}}$ in the question.
s is the distance traveled by the body under that acceleration which here is given as 80 cm i.e. 0.8 m.
v is the final velocity of the dumb-bell and u will be its initial velocity.
We will take u to be zero as the body starts from rest and we will take the downward direction as positive. So,
\begin{align} & 2\times 10\times 0.8={{v}^{2}}-{{0}^{2}} \\ & {{v}^{2}}=16 \\ & \therefore v=4m{{s}^{-1}} \\ \end{align}
The momentum of the body at that moment will be the product of its mass and velocity.
$P=mv=10\times 4=40kgm{{s}^{-1}}$
As all the momentum gets transferred, $40kgm{{s}^{-1}}$ will be transferred.

Note: The sign convention of the different quantities in the formula such as acceleration, displacement, and velocities used must be consistent with each other, positive for them must either be taken in the upwards direction or the downwards direction. We can also use the direct formula for the velocity of an object falling under gravity from rest which is $v=\sqrt{2gh}$.