Question

A bullet of 20 gm is fired in a sand filled bucket at a speed of 10m/sec and comes to rest in 5 sec find retarding force​

Answer 1

In the given problem,

mg=20kg ; vg=400m/s

mB=30g=30×10-3kg

no. of bullets fired/sec=4



For t=1sec.

Total mass of bullets in 1second=4×30=120g=0.120kg

Answer 2

Answer:

0.04 Newton.

Explanation:

$\underline{\bigstar\:\textsf{Given:}}$

• Mass (m) = 20 g = 20/1000 = 0.02 kg
• Initial velocity (u) = 10 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 5 s

$\underline{\bigstar\:\textsf{To\:find:}}$

• Retarding force (-F) = ?

$\underline{\bigstar\:\textsf{Solution:}}$

By second law of motion we comes to know that force = mass × acceleration.

For this we need to find acceleration, we also know that acceleration = (v - u)/t.

So, the final formula of force is $\green{\sf{F\:=\:m\:\times\:(\dfrac{v\:-\:u}{t})}}$

$\:$

$\Longrightarrow\:\sf{F\:=\:0.02\:\times\:(\dfrac{0\:-\:10}{5})}$

$\:$

$\Longrightarrow\:\sf{F\:=\:0.02\:\times\:\cancel{\dfrac{-10}{5}}}$

$\:$

$\Longrightarrow$ F = 0.02 × -2

$\:$

$\Longrightarrow$ F = -0.04 N

$\:$

$\Longrightarrow\sf{\underline{\underline{\therefore\:Retarding\:force\:=\:0.04\:N}}}$