Question

A bullet of 20 gram strikes a sandbag at a speed of 200 metre per second and gets embedded after travelling 2CM then find the resistive force exerted by sand on the bullet​

Answer 1

Answer:

20000N

Explanation:

Given-----> Mass of bullet is 20grams and bullet strikes a sandbag at a speed of 200 m/sec and gets embedded after travelling 2 Cm

To find -----> Find the resistive force exerted by the sand on the bullet to come to rest

Solution-----> ATQ,

Bullet of mass = 20 gram

= 20 × 10⁻³ kg

Initial velocoity of bullet = 200 m / sec

Final velocity of bullet = 0

Distance travelled by bullet = 2 Cm

= 2 × 10⁻² m

Now by third equation of motion ,

2as = v² - u²

=> 2 a ( 2 × 10⁻² ) = ( 0 )² - ( 200)²

=> 4 × 10⁻² a = - 40000

=> a = - 40000 / 4× 10⁻²

=> a = - 40000 × 10² / 4

=> a = - 10000 × 10²

=> a = - 10^6 m / s²

Negative value of a shows that , it is retardition and sand applied resistance at bullet ,

Now we know that ,

Force = mass × acceleration

So,

Resistive force applied by sand on bullet

= mass of bullet × acceleration

= 20 × 10⁻³ × 10^6

= 20× 10³

= 20000N

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Answer 2

Explanation:

mass=20g

$= \frac{20}{1000} kg$

=0.02kg

u=200m/s

v=0m/s

s=2cm

$= \frac{2}{100} m$

=0.02m

${v}^{2} = {u}^{2} +2as$

${0}^{2} = {200}^{2} + 2 \times a \times 0.02$

${- 200}^{2} = 2 \times a \times \frac{2}{100}$

$40000 = \frac{4a}{100}$

$a = 1000000 \: \frac{m}{ {s}^{2} }$

$f = ma$

$f = 1000000 \times 0.02$

$f = 10000 \times 2$

F=20000N