A 8.0×10−2 kg arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping.

What average force did the target exert on the arrow?
What average force did the arrow exert on the target?

Vf^2=Vi^2+2ad

Vf=0, Vi given, d given, solve for a.

Force=m*a

that formula gets me a wrong answer...

To find the average force exerted by the target on the arrow, we can use the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it.

The impulse is calculated using the formula:

Impulse = Change in momentum

Now, let's calculate the change in momentum of the arrow. The initial momentum (p1) is given by the mass (m) of the arrow multiplied by its initial velocity (v1):

p1 = m * v1

The final momentum (p2) is given by the mass (m) of the arrow multiplied by its final velocity (v2), which is zero (since the arrow stops):

p2 = m * v2 = 0

The change in momentum is given by:
Change in momentum = p2 - p1 = -p1

Now, we can substitute the given values into the equation. The mass of the arrow (m) is 8.0×10−2 kg, and its initial velocity (v1) is 25 m/s:

Change in momentum = - (m * v1) = - (8.0×10−2 kg * 25 m/s)

Calculating the change in momentum gives us the value.

To calculate the average force exerted by the target on the arrow, we now use the formula:

Average force = Impulse / Time

Since the time is not given, we are unable to calculate the average force exerted by the target on the arrow without additional information.

Similarly, to find the average force exerted by the arrow on the target, we would need additional information about the time taken for the arrow to penetrate the target.