You shoot an arrow at a block of wood. It hits and sticks into the block. Assuming the block of wood is on a frictionless surface and no energy is lost in the collision, at what velocity does the block of wood travel after the arrow hits?
Additional information for the problem:
m1 = 0.250 kg [the mass of the arrow]
v1 = 10.0 m/s [the velocity of the arrow moving away from you]
m2 = 5.00 kg [the mass of the block of wood]
v2 = 0 m/s [the velocity of the block]
There are two ways to do this problem. You can use either kinetic energy or momentum.
#1 using kinetic energy
Equations:
KE = ½mv2
Variables:
mf = 5.25 kg [the combined mass of the arrow and the block of wood]
vf = ? [the velocity of the arrow and the block, what we want to find]
E1 = ½m1v12 [the kinetic energy of the arrow before impact]
E2 = ½m2v22 [the kinetic energy of the block before impact]
E3 = ½mfvf2 [the kinetic energy of the arrow and the block after the impact]
We know that the energy of the system before and after are equal, so:
E1 + E2 = E3
½m1v12 + ½m2v22 = E3
½m1v12 + ½m2(0)2 = E3
½m1v12 + 0 = E3
½m1v12 = ½mfvf2
(m1v12)/(mf) = vf2
[(m1v12)/(mf)]0.5 = vf
Now we plug-in the given values for the variables.
[((0.250 kg)(10.0 m/s)2)/(5.25 kg)]0.5 = vf
2.18 m/s » vf
vf» 2.18 m/s away from you