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The momentum a dumb-bell of mass $10 kg$ transfers to the floor if it falls from a height of $80 cm$ is

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45 kg m / s

42 kg m / s

40 kg m / s

50 kg m / s

answer is C.

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Detailed Solution

The momentum which the

10 Kg

dumb-bell will transfer to the floor if it falls from a height of

80 cm

40 kg m / s .

Based on the given conditions:
Dumb-bell’s mass

(m)

= 10 kg

Height

(h) = 80 cm

\Rightarrow = \frac{80}{100} m

(As

1 m = 100 cm

)

\Rightarrow = 0.8 m

Initial velocity (u)

= 0 m / s

Final velocity

= v

Acceleration due to gravity

(g) = 10 m / s^{2}

Putting these values in the equation

v^{2} - u^{2} = 2 gh

v^{2} - (0) = 2 \times 10 \times 0.8 m

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

\Rightarrow v = {4 m / s}^{}

So, the final velocity is

4 m / s .

The law of conservation of momentum states “the momentum of a system is constant if no external forces are acting on the system”.
We know, momentum

p = m \times v

\Rightarrow p = 10 kg \times 4 m / s

\Rightarrow p = 40 kg m / s

By the law of conservation of momentum, this entire momentum will be transferred to the floor.

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