A golf club colliding with a stationary ball?

Question

A golf putter (for the sake of the question) is moving at X velocity and strikes a stationary golf ball....what will the velocity be of the golf ball after the collision.

Putter mass: .133 kg
Putter velocity: 1.25 m/s
Angle of the face of the putter: 3 degrees
Golf ball mass: .0047 kg

I think this is a conservation of momentum problem, but does anyone know how to incorporate the 3 degree angle of the putter face too?

Thanks....best answer will be the one that shows the most work.

the putter head continues (albeit at a slower velocity after the impact)...the putter head velocity after the impact is also unknown (this is a real world problem and not from some class or text book)

also, the putter head continues in the same direction after it collides with the ball....just as would happen in real life disregarding any miniscule angular differences

Answer 1

You still need the velocity (vector) of the putter after the impact, or velocity of the ball (since that's what we're looking for, we won't consider we have it).

Since no information was given to calculate potential energy (PE) absorbed by the ball (more realistic), I'm going to assume an inelastic collision.
KEi = KEf or (m1)(v1i) + (m2)(v2i) = (m1)(v1f) + (m2)(v2f)
where m = mass, v = velocity, i = initial, f = final, 1 = club face and 2 = ball

The ball initial velocity is 0, so the resultant formula is
((m1)(v1i) = (m1)(v1f) + (m2)(v2f).

At this point, you have to stop. You have one equation, and two unknowns (v1f and v2f). You need to know how fast the putter is going after the impact.

If you assume that the putter head takes no deflection in the verticle direction (from the impact with the ball) this simplifies things, becasue the y component falls out completely (in effect the angle of the face of the putter becomes unimportant), and leaves you with:

(v2f) = (35.4) - (28.3)(v1f)

You still need the final velocity of putter head (v1f) after the collision.

If you don't assume this, you have to break down into components:
X component:
(m1)(v1ix) = (m1)(v1fx)(cos(angle of deflection of putter)) + (m2)(v2f)(cos(3 deg))

(.133kg)(1.25m/s) = (.133kg)(v1f)(cos(angle of deflection of putter)) + (.0047kg)(v2f)(.998)

Y component:
(m1)(v1iy) = 0 = (m1)(v1f)(sin(angle of deflection of putter)) + (m2)(v2f)(sin(3 deg))

0 = (.133kg)(v1f))(sin(angle of deflection of putter)) + (.0047kg)(.05)(v2f)

You still need the final velocity of putter head (v1f) after the collision, but as a vector (which includes the angle of deflection), so you can solve the equations.

Ouch, I think I hurt my brain.

Note: As an example, I assumed no verticle deflection of the putter head, and a final putter speed of 1.2 m/s. This resulted in a ball speed of 1.4m/s, or approximately 4.5 yards/second. That seemed a reasonable speed, though I could not find anything about putting speed on the internet. The net effect is that a little slowing of the putter head results in a lot of speed on the ball.

Answer 2

In real life, you can't solve this problem. You have to consider whether there is acceleration through the ball, whether energy is lost through compression of the ball, whether you hit the ball in the center or if it hops in the air. Just because the putter has 3 degrees doesn't mean that is how you make contact, whether you putter stays in a horizontal plane or makes a U, and when it comes to a stop. So there is both conservation of mom and conservation of energy.

Answer 3

♠ putter does not matter after collision. The putter’s velocity has 2 components: 1 is normal to the face and transferring momentum to the ball, another 1 is parallel to the face and not participating in our business. Draw a pic!
♣ momentum conservation law: 0.133*1.25*cos3° = 0.0047*|v|, hence
|v|= 0.133*1.25*cos3° /0.0047 =35.32 m/s
♦ let x-axis be the direction of putter’s velocity, then v=35.32(cos3°*i +sin3°*j), v being normal to putter’s face;