Impulse & Momentum: Golf Ball & Club Force Calcs

[elimenohpee]

Homework Statement

A golfer hits a golf ball of mass 0.045 kg the ball over some short trees. He hits the ball at
an angle of 60(degrees) to the horizontal and it travels a horizontal distance (Range) of 60.0 m in a time of 8.00 s. The golf club of mass 0.60 kg is in contact with the ball for a time of 2.40 ms.
(a) What is the average impulsive force on the golf ball?
(b) What is the average impulsive force on the golf club?
(c) What is the change in momentum of the golf club?

Homework Equations

All my answers are exactly have of what my teacher posted. My answer for (a) should be 280N, (b) should be -280N, and (c) should be -0.67 kg m/s

The Attempt at a Solution

Since the ball is hit at an angle, I split the momentum into x and y components. Initial velocity is zero, so both x1 and y1 are zero.
(a) The final momentum for x would be: mvcos(theta) = (0.045kg)(7.5m/s)(cos(60))= 0.169 kg m/s

The final momentum for y would be: mvsin(theta) = (0.045kg)(7.5m/s)(sin(60))= 0.29 kg m/s
Divide both values by the time the force was in contact (2.5 ms).
So to find the average impulsive force, take both values of momentum divided by time and find the magnitude: [(70.4)^2 + (121)^2]^0.5 = 140 N

(b) it would just be the opposite of the force on the golf ball, -140N
(c) change in momentum would just take the magnitude of the the momentum:
[(0.169)^2 + (0.29)^2]^0.5 = -0.33 kg m/s (negative because its the change in momentum of the golf club not the golf ball)

 

[LowlyPion]

I might note that the 60m/8s is just the x component of velocity, not the initial x,y velocity of the ball.

 

[nlightner]

Your approach to solving this problem is correct. Your teacher's answers may be different due to rounding or slight variations in calculations. It is important to note that the average impulsive force is the same for both the golf ball and the golf club, but the direction is opposite. This is because of Newton's third law of motion - for every action, there is an equal and opposite reaction. Good job on correctly using the equations for momentum and understanding the concept of impulse and momentum!