Question

Please write legibly and write out all equations and units.

Please show all steps. Algebra-based physics only please.

17. A 0.02 kg bullet is shot into an 3.0 kg block of wood suspended from a 0.5 m long string. The bullet becomes imbedded in the block of wood, and the block (with the bullet inside) swings up, as shown in the figure. If the speed of the bullet just before impact is 300 m/s, how high vertically does the block (with the bullet inside) swing up into the air? Neglect air resistance. v 300 m/s Th ? 0.20 m 0.42 m В 0.46 m C 0.30 m 0.19 m Е 0.5 т

Answer 1

1st use conservation of momentum

Mb * Vb + Mw * Vw = ( Mb + Mw) * V

0.02*300 + 3.0 * 0 = 3.02 * V

V = 1.986 m/sec

this is velocity of with bullet just after bullet hit the wood block

we will use this point ( when bullet is just attached to block ) and the highest point ( where block velocity is zero ) to use the conservation of energy principal.

now at maximum height velocity of block would be zero

1/2 mv^2 = mg*h

or h = v^2/(2*g) = 3.947/(2*9.81) = 0.201 m answer

feel free to drop comment if you have any doubt.