Answer:
4.79m/s
Step-by-step explanation:
According to law of conservation of momentum;
The sum of momentum of the bodies before collision is equal to the momentum after collision.
m1u1 + m2u2 = (m1+m2)v
Given;
m1 = 0.3kg
u1 = 2.30m/s
m2 = 0.0225kg
u2 = 38m/s
Required
speed of the arrow after passing through the target v
Substituting the given data into the formula
0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v
0.69 + 0.855 = 0.3225v
1.545 = 0.3225v
v = 1.545/0.3225
v = 4.79m/s
Hence the speed of the arrow after passing through the target is 4.79m/s