Question

A bullet of mass 40 g moving with a speed of 90 ms" enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistive force exerted by the block on the bullet is : (1) 180 N (2) 220 N (3) 270 N (4) 320 N

Answer 1

Given:-

Mass of bullet(m)=40g

=>40/1000

=>0.04kg

Initial velocity(u)=90m/s

Final velocity(v)=0 (as it finally comes to rest)

Distance(s)=60cm

=>0.6m

To find:-

Average resistive force exerted by block on the bullet(F)

Solution:-

By,using the third equation of motion,we get-----

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

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

$= > - 8100 = 1.2a$

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

$= > a = - 6750ms ^{ - 2}$

Now,we know that

F=ma

F= 0.04×(-6750)

F= -270N

Thus,the resistive force is -270N or 270N in the opposite direction.

Hence,the correct option is (3)270N

Answer 2

Answer:

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