Question

A bullet of mass 200g moving with a speed of 80m/s enters a cement wall and is stopped after a distance of 40cm. Find the average resistive force exerted by wall on bullet
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Answer 1

Answer:

Mass of bullet(m)= 30g = 0.03kg.

Initial velocity (u)= 400m/s.

Final velocity (v)= 0m/s. [Since bullet comes to rest after penetrating 10 cm in wall]

Distance penetrated by bullet (S) or displacement of bullet in presence of resistive force = 10 cm = 0.1m.

Acceleration (a)=?

From third equation of motion:

v^2=u^2+2aS

0^2= 400^2+2a*0.1

0.2a=-160000

a=-160000/0.2= -8,00,000m/s^2.

The negative sign shows that acceleration takes place in direction opposite to the initial direction of motion of bullet or retardation takes place.

Avg. force (f)= m*a

f= 0.03*-800000

f= -24000N

Here too negative sign shows that force applied by mud wall on bullet is in direction opposite to initial direction of motion of bullet.

Answer 2

Answer:3200 N

Explanation:

1/2m(v1^2−v2^2)=F⋅x

or, F=$\frac{m(v_{1} ^{2} -v_{2} ^{2}) }{2x}$

or, F= $\frac{0.2*80*80}{0.4}$

or, F= 80*40

or, F=3200 N

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