Question

a bullet of mass 0.04kg moving with a speed of 90m/s enters a heavy wooden block and is stopped after a distance of 60cm what is the average resistive force exerted by the block on the bullet

Answer 1

Given :

• mass = m= 0.04kg

• initial velocity = u=90m/s

• final velocity = v= 0m/s

• displacement = s= 60cm =0.6m

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Need to Find:

• The Force acting on body=?

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Solution :

By 3rd equation of motion,

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

$\implies \: {0}^{2} - {90}^{2} = 2a \times 0.6$

$\implies \: - 8100 = 0.6 \times 2 \times a$

$\implies \: a = \dfrac{ - 81000}{2 \times 6}$

$\pink{\boxed{\boxed{\bold{ \implies a= 6750 m/ s^{2}}}}}$

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From NLM, We know,

F= m× a

$\implies F= 0.04 \times (-6750)N$

$\red{\bold{\boxed{\large{\implies Force= (-270)N}}}}$

Here the minus (-) symbol indicates that the Force of 270N is acting in opposite to the direction of motion.

Answer 2

The average resistive force exerted by the block on the bullet is -270N

Given :-

Mass of the bullet (m) = 0.04kg

Initial velocity of the bullet (u) = 90m/s

Final velocity of the bullet (v) = 0m/s

Displacement taken by the bullet (s) = 60cm

(1m = 100cm) = 0.6m

To Find :-

Force exerted by the block on the bullet

Formula Applied :-

III equation of motion

$\frac{ {v}^{2} - {u}^{2}}{2 \times s} = a$

Newton's II law of motion,

$force = mass \times acceleration$

$f = m \times a$

Solution :-

By third equation of motion ,

${v}^{2} - {u}^{2} = 2as \\ \frac{ {v}^{2} - {u}^{2} }{2 \times s} = a$

$a = \frac{ {0}^{2} - {90}^{2} }{2 \times 0.6}$

$a = \frac{ - 8100}{1.2}$

$a = - 6750$

Hence the acceleration of the bullet is -6750m/s

Now, by using newton's second law of motion by which

Force = Mass x Acceleration

$f = m \times a$

$f = 0.04 \times - 6750$

$f = - 270N$

The negative sign indicates that the force of (-270N) is acting in the opposite direction of the motion