Question

a bullet of mass 20 gram is moving with a velocity of 100m/s strikes a wooden block of mass 80 g and gets embedded in it.calculate the velocity of combined system​

Answer 1

Answer:

Given Mass of bullet = 20gm

Mass of wooden block = 980gm

initial velocity = 20m/s

Final velocity = 0m/s (as it got embedded in wooden block)

Initial velocity of wooden block = 0 m/s

Final velocity = v

So from conservation of momentum

m1(0-20)/t = m2(v-0)/t

20(-20) = 980(v)

v = -400/980 = -20/49= -0.41m/s

so velocity of block is 0.41m/s

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Explanation:

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Answer 2

Explanation:

let bullet be object 1 and let wooden block be object 2

object 1:

m1-20 gram = 0.002 kg

u1- 100m/s

v1- 0

now, object 2:

m2-80 gram : 0.008 kg

u2: 0

V2:?

acc. to law of conservation of momentum

m1u1+ m2u2 = m1v1 + m2v2

0.002*100 + 0.008*0 = 0.002*0 + 0.008*V2

0.2 + 0 = 0+ 0.008*V2

0.2 = 0.008* V2

0.2/ 0.008 = V2

V2 = 25 m/s

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