High-speed stroboscopic photographs show that the head of a golf club of mass 230g is traveling at 56.1m/s just before it strikes a 44.6g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 41.4m/s. Calculate the speed of the golf ball just after impact
See previous post: 10-24-17, 6:05 PM.
To calculate the speed of the golf ball just after impact, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is defined as the product of its mass and velocity. Mathematically, momentum (p) is given by the equation p = m * v, where m is the mass of the object and v is its velocity.
Before the collision, the golf club head is traveling at 56.1 m/s, and the golf ball is at rest with a velocity of 0 m/s. Therefore, the initial momentum of the golf club head is given by p_initial_club = m_club * v_initial_club = (0.230 kg) * (56.1 m/s).
After the collision, the golf club head is traveling at 41.4 m/s, and the golf ball has a velocity of v_ball. The mass of the golf ball is 0.0446 kg. The final momentum of the golf club head is given by p_final_club = m_club * v_final_club = (0.230 kg) * (41.4 m/s). The final momentum of the golf ball is given by p_final_ball = m_ball * v_final_ball = (0.0446 kg) * v_ball.
Using the principle of conservation of momentum, we can set up the equation:
p_initial_club = p_final_club + p_final_ball
Substituting the values we know, we have:
(0.230 kg) * (56.1 m/s) = (0.230 kg) * (41.4 m/s) + (0.0446 kg) * v_ball
Simplifying, we have:
12.903 kg * m/s = 9.534 kg * m/s + (0.0446 kg) * v_ball
Now, let's solve for v_ball:
12.903 kg * m/s - 9.534 kg * m/s = (0.0446 kg) * v_ball
3.369 kg * m/s = (0.0446 kg) * v_ball
Divide both sides of the equation by 0.0446 kg:
(3.369 kg * m/s) / (0.0446 kg) = v_ball
v_ball = 75.671 m/s
Therefore, the speed of the golf ball just after impact is approximately 75.671 m/s.