High-speed stroboscopic photographs show
that the head of a 198 g golf club is traveling at 24.4 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 17.3 m/s.
Find the speed of the golf ball immediately
after impact.
Answer in units of m/s
conserve momentum
0.198 * 24.4 = 0.198 * 17.3 + 0.0459 v
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
The momentum before the collision is given by the product of the mass and velocity of the golf club head:
Momentum before = mass of golf club head * velocity of golf club head before collision
Momentum before = 198 g * 24.4 m/s
Next, we need to calculate the momentum after the collision. After the collision, the golf club head is traveling at 17.3 m/s. The golf ball, initially at rest, gains some velocity and moves in the same direction as the golf club head.
Let's assume the velocity of the golf ball after the impact is represented by v.
Therefore, the momentum after the collision is given by the sum of the momentum of the golf club head and the momentum of the golf ball:
Momentum after = (mass of golf club head * velocity of golf club head after collision) + (mass of golf ball * velocity of golf ball after collision)
Momentum after = (198 g * 17.3 m/s) + (45.9 g * v)
Since momentum is conserved, we can equate the momentum before the collision to the momentum after the collision:
Momentum before = Momentum after
198 g * 24.4 m/s = (198 g * 17.3 m/s) + (45.9 g * v)
Now we can solve for v, the velocity of the golf ball after the impact:
(198 g * 24.4 m/s) - (198 g * 17.3 m/s) = 45.9 g * v
v = ((198 g * 24.4 m/s) - (198 g * 17.3 m/s)) / (45.9 g)
v ≈ 5.80 m/s
Therefore, the speed of the golf ball immediately after impact is approximately 5.80 m/s.
To solve this problem, we can apply the principles of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision.
Let's assume that the golf ball's speed after the impact is v_ball.
Before the collision:
- The mass of the golf club (m_club) is 198 g, and its speed (v_club) is 24.4 m/s.
- The mass of the golf ball (m_ball) is 45.9 g, and its initial speed is 0 m/s (at rest).
After the collision:
- The speed of the golf club (v_club) is given as 17.3 m/s.
- The speed of the golf ball (v_ball) is what we need to find.
According to the principle of conservation of momentum:
(m_club * v_club) + (m_ball * 0) = (m_club * v_club) + (m_ball * v_ball)
Let's plug in the values and solve for v_ball:
(0.198 kg * 24.4 m/s) + (0.0459 kg * 0 m/s) = (0.198 kg * 17.3 m/s) + (0.0459 kg * v_ball)
4.8312 kg·m/s = 3.4294 kg·m/s + (0.0459 kg * v_ball)
Now, let's isolate v_ball:
4.8312 kg·m/s - 3.4294 kg·m/s = 0.0459 kg * v_ball
1.4018 kg·m/s = 0.0459 kg * v_ball
Divide both sides by 0.0459 kg:
v_ball = 1.4018 kg·m/s / 0.0459 kg
v_ball ≈ 30.49 m/s
Therefore, the speed of the golf ball immediately after impact is approximately 30.49 m/s.