A golf club exerts an average horizontal force of 1000 n on a 0.045 -kg golf ball that is initially at rest on the tee. the club
1 answer:
The impulse (the variation of momentum of the ball) is related to the force applied by
where
is the variation of momentum, F is the intensity of the force and 
is the time of application of the force.
Using F=1000 N and
, we can find the variation of momentum:
This can be rewritten as
where
and 
are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
from which we find the final velocity of the ball:
where
Using F=1000 N and
This
where
from which we find the final velocity of the ball:
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Answer:
3.16 m·s⁻¹ at an angle of 71.6°
Explanation:
Assume that the diagram is like Fig. 1 below.
The boat is heading straight across the river and the current is directed straight downstream.
We have two vectors at right angles to each other.
1. Calculate the magnitude of the resultant
We can use the Pythagorean theorem (Fig. 2).
R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²
R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹
2. Calculate the direction of the resultant
The direction of the resultant is the counterclockwise angle (θ) that it makes with due East .
tanθ = opposite/adjacent = 3/1 = 3
θ = arctan 3 = 71.6°
To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.
