Average force necessary to stop a hammer having momentum 50, N-S in 0.25, sec ...... N

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4-1.Newton's Laws of Motion

easy

The average force necessary to stop a hammer having momentum $50\, N-S$ in $0.25\, \sec$ ...... $N$

A

$50$

B

$12.5$

C

$100$

D

$200$

Solution

$\Delta p=F_{a v e} \cdot \Delta t$

$\Rightarrow \mathrm{F}_{\mathrm{ave}}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$

$=\frac{50}{0.25}=200 \mathrm{N}$

Standard 11

Physics

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