An archer released a 100 g arrow at a velocity of 20 m/sec. It hits its target having a mass of 600 g. With the arrow embedded in the target, with what velocity does the target move?
Answer (1)
To find the velocity of the target after the arrow has hit and embedded in it, we can use the principle of conservation of momentum. This principle states that if no external forces are acting on a system, the total momentum of that system before an event must be equal to the total momentum of that system after the event.
Let's denote:
The mass of the arrow, m 1 = 100 g = 0.1 kg .
The velocity of the arrow before it hits the target, v 1 = 20 m/s .
The mass of the target, m 2 = 600 g = 0.6 kg .
The initial velocity of the target, v 2 = 0 m/s (it is stationary).
The final velocity of the combined arrow and target after impact, v f .
Using the conservation of momentum:
m 1 ⋅ v 1 + m 2 ⋅ v 2 = ( m 1 + m 2 ) ⋅ v f
Substituting the known values:
0.1 ⋅ 20 + 0.6 ⋅ 0 = ( 0.1 + 0.6 ) ⋅ v f
2 = 0.7 ⋅ v f
To solve for v f , divide both sides by 0.7:
v f = 0.7 2 ≈ 2.86 m/s
Therefore, the velocity of the target, with the arrow embedded in it, is approximately 2.86 m/s .
