A bullet with a mass of 0.01kg is fired horizontally into a block of wood hanging on a string.The bullet stick

Question

in the wood and causes it to swing upward to a height of 0.1m. If the mass of the wood block is 2kg, What was the initial speed of the bullet?

Answer 1

Conservation of energy: KE=PE
1/2*Mb*Vb^2= (Mb+Mw)*g*h
Vb^2=2*(Mb+Mw)*g*h/Mb
Vb^2=2*(0.01+2)*10*0.1/0.01
Vb^2=402
Vb=20.05 m/s

Answer 2

1) use conservation of momentum to solve for the final speed with the bullet embedded in the block:

m(bullet) x v0(bullet) + M(block) x v0(block) = Vf x (m(bullet) + M(block)}

because the block is not moving:

mv0 + 0 = Vf (m+M) then: Vf = mv0/(m+M)

2) then when the bullet is embedded in the block, they travel together until it reaches a height of 0.1 (m), so Kinetic energy is converted to Potential Energy:

KE = PE

I/2 x (m+M) x (Vf)^2 = (m+M) x g x h

(m+M) cancels, multiply 2 over to other side then:

(Vf)^2 = 2gh

take the Vf we found and square it, move the variables and you get:

V0(bullet) = {RAD[2gh(m^2 + 2mM + M^2]}/m (meters/second)

Whew.......

Answer 3

Calculate the potential energy contained in the wood block *plus* the mass of the bullet (2.01 kg total) as mgh and set it equal to the kinetic energy of the bullet (which is .5mv²) then solve for v.


Doug

Answer 4

enable the fee of the bullet grew to become into u m/s & the fee of block+bullet grew to become into v m/s, =>via the regulation of capacity conservation:- =>KE(preliminary) = PE(very final) =>one million/2(M+m)v^2 = (M+m)gh =>v = ?2gh =>v = ?[2 x 9.8 x 15 x 10^-2] =>v = one million.seventy one m/s via the regulation of momentum conservation:- =>mu = (M+m)v =>15 x 10^-3 x u = (15 x 10^-3 + 2.5) x one million.seventy one =>u = 287.40 9 m/s

Answer 5

20 metres per second