Question

A bullet of 10g moving with velocity 400moving with velocity of 400 m/s gets embedded in afreely suspended wooden block of mass 90 g what is the velocity acquired by the block pldsss answer fast

Answer 1

Let the final velocity of the bullet and the wooden block be= V m/s

So we have the data:-

Mass of bullet(m1) =10g=0.01kg

Initial velocity of bullet(u1) =400m/s

Mass of block(m1) =90g=0.09kg

Initial velocity of block(u2) =0m/s

Final velocity of bullet(v1) = Final velocity of block(v2) = V m/s

As per Newton's third law of motion :-

m1.u1 + m2.u2 = m1.v1 + m2.v2

=>m1.u1 + m2.u2 = (m1 + m2) V

(As v1=v2=V)

=>V = m1.u1 + m2.u2/ m1 + m2

V = 0.01×400 + 0.09×0/ 0.1 +0.9

V = 4 + 0 /1

➡️V = 4 m/s

The block gains a velocity of 4 m/s

Hope this helps you