Question

A SHOT OF MASS 0.5KG MOVING HORIZONTALLY AT 0.6M/S IS STOPPED BY 0.1M OF SAND. WHAT AVERAGE FORCE OF RESISTANCE IS EXERTED BY THE SAND ON THE SHOT ? WHAT IS THE AVERAGE RETARDATION?\ URGENT ANSWER

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.

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.

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