Question

A bullet of mass 50g strikes a wooden block with speed 1000m/s and penetrates up to a distance of 5 cm.calculate the resistive force exerted by the wooden block on bullet and the time taken by the bullet to come to rest​

Answer 1

Given:-

• Mass of bullet(m) is 50g = 0.05kg
• Speed of bullet(u) is 1000m/s
• Distance(s) = 5cm = 0.05m

To find:-

• The resistive force exerted by wooden block on bullet.
• And the time taken by the bullet to come to rest.

Solution :-

✏ We need to find force, and force is equal to multiples of mass and acceleration, F = ma, so first we've to calculate acceleration of the body.

From third equation of motion,

$\sf\implies \: 2as = v {}^{2} - u {}^{2} \\ \sf\implies \:2 \times a \times 0.05 = (0) {}^{2} - (1000) {}^{2} \\\sf \implies \:0.1a = - 1000000 \\\sf \implies \: a = \frac{ - 1000000}{0.1} ms {}^{2} \\ \sf \implies\boxed{ \bf{a = -10000000ms {}^{2} }}$

✏ Using Newton's second law, force will be calculated as,

$\sf \implies \: F= ma \\ \sf \implies \: F = (0.05kg)( - 10000000ms {}^{2} ) \\ \sf \implies \boxed{ \boxed{ \bf{ \purple{F = - 500000N}}}}$

∴ the resistive force is -500000N which is exerted by the wooden block on bullet.

✏ Now we need to find time taken by the bullet to come to rest.

By using first equation of motion,

$\sf \implies \: v = u + at \\ \sf \implies \: t = \dfrac{v - u}{a} \\ \sf \implies \: t = \frac{0 - 1000}{ - 10000000} \\ \sf \implies \small{\boxed{\boxed{ \bf{ \purple{t = 0.0001 \: seconds}}}}}$

∴ time taken by the bullet to come to rest is 0.0001 sec.

Extra infro:-

• First equation of motion => (v = u+at) [used in 3rd part]

• Newton's second law of motion => (F = ma) [used in 2nd part]

• Third equation of motion => (2as = v²-u²) [used in 1st part]

Answer 2

Answer:-

➣

• Resistive Force exerted by the wooden block on bullet is $\bf 500 \: kilo Newton [kN]$

• The time taken by the bullet to come to rest is $\bf 0.1 \: milli seconds$

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• Given:-

Mass of bullet [m] = $\bf{50 \: g}$

$\implies {\dfrac{50}{1000}}kg$

$\implies\bf{0.05 \: kg}$

Initial speed of bullet [u] = $\bf{1000 \: m/s}$

Distance covered [s] = $\bf{5 \: cm}$

$\implies{\dfrac{5}{100}}m$

$\implies\bf{0.05 \: m}$

Final velocity of bullet [v] = $\bf{0}$

• To Find:-

Resistive force exerted by wooden block on bullet [F] = $\bf{?}$

Time taken by the bullet to come to rest [t] = $\bf{?}$

Formula Used:

Third equation of motion:-

$\boxed{v^2-u^2=2\;a\;s}$

Newton's second Law of motion:-

$\boxed{F=m\;a}$

First equation of motion:-

$\boxed{v=u+a\;t}$

• Solution:-

$\bf{\mapsto}$Firstly, Calculating acceleration of the bullet,using third equation of motion:-

$\bf\implies{v^2-u^2=2\;a\;s}$

$\longrightarrow{(0)^2-(1000)^2=2\;a\;(0.05)}$

$\longrightarrow{-1000000=0.1\;a}$

$\large\implies\bf\boxed{a=-10000000\;ms^{-2}}$

$\bf{\mapsto}$Now, calculating Force exerted by the bullet on the wooden block,using Newton's second law of motion:-

$\bf\implies{F=m\;a}$

$\longrightarrow{F=(0.05)\times(-10000000)}$

$\large\implies\bf\boxed{F=-500000\;N}$

$\bf\large{OR}$

$\large\implies\bf\boxed{F= - 500 kN}$

• Since, The bullet came to rest on collision with the wooden block hence, the Force exerted by Wooden block on the bullet will be $\bf{=}$to the force exerted by bullet on wooden block.

Therefore,

• Force exerted by wooden block on bullet is $\bf{500 \: kilo Newton[kN]}$. [In the direction opposite to motion of bullet].

$\bf{\mapsto}$Lastly,calculating time taken by bullet to come to rest,using first equation of motion:-

$\bf\implies{v=u+a\;t}$

$\longrightarrow{0=1000+(-10000000)\;t}$

$\longrightarrow{-1000=-10000000\;\;t}$

$\large\implies\bf\boxed{t=0.0001\;sec}$

$\large\implies\bf\boxed{t = 0.1 \: milli \: second}$

Therefore,

Time taken by bullet to come to rest is $\bf{0.1 \: milli seconds}$

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