Questions: When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 43.6 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 25.0 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.

Solution Steps

We are given a problem involving a collision between a golf club head and a golf ball. The mass of the club head is \(158 \, \text{g}\) (or \(0.158 \, \text{kg}\)) and its initial velocity is \(43.6 \, \text{m/s}\). The mass of the golf ball is \(46.0 \, \text{g}\) (or \(0.0460 \, \text{kg}\)) and it is initially at rest. After the collision, the club head's velocity is \(25.0 \, \text{m/s}\). We need to find the velocity of the golf ball immediately after the collision.

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. The equation for conservation of momentum is:

\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]

where:

• \(m_1 = 0.158 \, \text{kg}\) is the mass of the club head,
• \(v_{1i} = 43.6 \, \text{m/s}\) is the initial velocity of the club head,
• \(v_{1f} = 25.0 \, \text{m/s}\) is the final velocity of the club head,
• \(m_2 = 0.0460 \, \text{kg}\) is the mass of the golf ball,
• \(v_{2i} = 0 \, \text{m/s}\) is the initial velocity of the golf ball,
• \(v_{2f}\) is the final velocity of the golf ball, which we need to find.

Substitute the known values into the conservation of momentum equation:

\[ (0.158 \, \text{kg})(43.6 \, \text{m/s}) + (0.0460 \, \text{kg})(0 \, \text{m/s}) = (0.158 \, \text{kg})(25.0 \, \text{m/s}) + (0.0460 \, \text{kg})v_{2f} \]

Simplify and solve for \(v_{2f}\):

\[ 6.8888 \, \text{kg} \cdot \text{m/s} = 3.95 \, \text{kg} \cdot \text{m/s} + 0.0460 \, \text{kg} \cdot v_{2f} \]

\[ 6.8888 \, \text{kg} \cdot \text{m/s} - 3.95 \, \text{kg} \cdot \text{m/s} = 0.0460 \, \text{kg} \cdot v_{2f} \]

\[ 2.9388 \, \text{kg} \cdot \text{m/s} = 0.0460 \, \text{kg} \cdot v_{2f} \]

\[ v_{2f} = \frac{2.9388 \, \text{kg} \cdot \text{m/s}}{0.0460 \, \text{kg}} \]

\[ v_{2f} = 63.8870 \, \text{m/s} \]

Final Answer