A 0.027 kg bullet is firedstraight up at a falling wooden block that has a mass of4.0 kg. The bullet has a speed of710 m/s when it strikes the block. Theblock originally was dropped from rest from the top of a buildingand had been falling for a time

s

*t*when the collision withthe bullet occured. As a result of the collision, the block (withthe bullet in it) reverses direction, rises, and comes to amomentary halt at the top of the building. Find the time*t*.s

## Answers ( 7 )

The velocity equation for the block dropped from rest for a timet is v

_{1B}= -gt. The mass of the block is m_{1}=4 kg, that of the bullet is m_{2}= 0.027 Kg, and thevelocity of the bullet before the collision is v_{2B}= 710m/s. Note that after the collision, the velocity of the block (nowwith the bullet in it) is exactly reversed to v_{f}=gtThis is a completely inelastic collision. Applyingconservation of momentum, we have

m

_{1}v_{1B}+ m_{2}v_{2B}=(m_{1}+ m_{2})v_{f}_{2}v_{2B}- m_{1}gt = (m_{1}+ m_{2})gt. Solving this for the time yeilds:_{2}v_{2B})/(2m_{1}+m_{2})g = 0.24369 s^{2}-v'^{2}= -2gs---1^{2}^{2}-v'^{2}= -2gs---1^{2}^{2}-v'^{2}= -2gs---1^{2}