A 12 g fly, attached to a fly fishing rod, comes undone and hits a wall with a velocity of 3.2 m/s to the right. It rebounds with a velocity of 1.2 m/s. If the collision last for 0.02 seconds, what is the magnitude of the resultant force fell on the fly?
To find the magnitude of the resultant force acting on the fly, we can apply Newton's second law, which states that the force acting on an object is equal to the rate of change of its momentum.
The momentum of an object can be calculated by multiplying its mass (m) with its velocity (v). So, the momentum before the collision (P1) is given by:
P1 = m * v1
Where:
m = mass of the fly = 12 g = 0.012 kg
v1 = velocity before the collision = 3.2 m/s
Therefore, P1 = 0.012 kg * 3.2 m/s = 0.0384 kg·m/s.
The momentum after the collision (P2) is given by:
P2 = m * v2
Where:
v2 = velocity after the collision = 1.2 m/s
Therefore, P2 = 0.012 kg * 1.2 m/s = 0.0144 kg·m/s.
Since the collision lasts for 0.02 seconds, we can calculate the change in momentum (ΔP) using the formula:
ΔP = P2 - P1
Therefore, ΔP = 0.0144 kg·m/s - 0.0384 kg·m/s = -0.024 kg·m/s.
The negative sign indicates that there is a change in the direction of momentum during the collision.
Finally, the magnitude of the resultant force (F) can be found by dividing the change in momentum by the duration of the collision:
F = ΔP / Δt
Where:
Δt = duration of the collision = 0.02 seconds
Therefore, F = -0.024 kg·m/s / 0.02 s = -1.2 N.
The negative sign indicates that the force is exerted in the opposite direction of the fly's initial motion, which in this case is to the left.
Hence, the magnitude of the resultant force applied to the fly during the collision is 1.2 Newtons.