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3 years ago

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A golfer hits a golf ball with a club head velocity of 94 mph. Mass of golf club head (m): 190 g Mass of golf ball (ms): 46 g Co

efficient of restitution (COR) (e): 0.83 What is most nearly the velocity of the golf ball after being hit?

1 answer:

MAXImum [283]3 years ago

7 0

Answer:

Velocity of golf ball≅ 12.87mph

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= $\frac{Vf2- Vf1}{Vi1 - Vi2}$

0.83= $\frac{Vf2 - Vf1}{94 - 0}$

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph

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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call $v_f$ the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
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$m_B=2 kg$ is the mass of ball B
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and the positive sign means the two balls are going to the right.

2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
$m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}$
$\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2$
where
$v_{fA}$ is the final velocity of ball A
$v_{fB}$ is the final velocity of ball B

If we solve simultaneously the two equations, we find:
$v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s$
$v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s$
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
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and the total momentum after the collision is:
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