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3 years ago

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What is the force of a lead ball that starts with a momentum of 7 kgm/s and end with a momentum of 22 kgm/s over a time of 31

seconds?
Explain how you got your answer.

1 answer:

Gwar [14]3 years ago

6 0

Answer:

F = 0.483 N

Explanation:

Initial momentum, $P_i= 7\ kg-m/s$

Final momentum, $P_f= 22\ kg-m/s$

Time, t = 31 s

We need to find the force of a lead ball. We can use here the impulse momentum theorem.

$\text{Change in momentum}=F\times t$

F is force

$P_f-P_i=Ft\\\\F=\dfrac{P_f-P_i}{t}\\\\F=\dfrac{22-7}{31}\\\\F=0.483\ N$

So, the force is 0.483 N.

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therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

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$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

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$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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