High-speed stroboscopic photographs show

that the head of a 197 g golf club is traveling at 36.8 m/s just before it strikes a 45.9 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 25.9 m/s. Find the speed of the golf ball immediately after impact.
Answer in units of m/s

To find the speed of the golf ball immediately after impact, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision, as long as no external forces are acting on the system.

Before the collision, the golf club head is traveling at 36.8 m/s, and its mass is 197 g (or 0.197 kg). The golf ball is at rest on a tee, so its initial velocity is 0 m/s, and its mass is 45.9 g (or 0.0459 kg).

After the collision, the golf club head is traveling at 25.9 m/s, and we need to find the final velocity of the golf ball.

Let's denote the initial velocity of the golf club head as v1_club, the final velocity of the golf club head as v2_club, the initial velocity of the golf ball as v1_ball, and the final velocity of the golf ball as v2_ball.

According to the conservation of momentum:
(m1 * v1_club) + (m2 * v1_ball) = (m1 * v2_club) + (m2 * v2_ball)

Plugging in the given values:
(0.197 kg * 36.8 m/s) + (0.0459 kg * 0 m/s) = (0.197 kg * 25.9 m/s) + (0.0459 kg * v2_ball)

Simplifying the equation:
7.2436 kg*m/s = (5.1123 kg*m/s) + (0.0459 kg * v2_ball)

Now, let's isolate v2_ball:
7.2436 kg*m/s - 5.1123 kg*m/s = 0.0459 kg * v2_ball
2.1313 kg*m/s = 0.0459 kg * v2_ball

Dividing both sides of the equation by 0.0459 kg:
(2.1313 kg*m/s) / 0.0459 kg = v2_ball
46.5018 m/s = v2_ball

Therefore, the speed of the golf ball immediately after the impact is approximately 46.5018 m/s.