High-speed stroboscopic photographs show
that the head of a 212 g golf club is trav-eling at 29.8 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 21.8 m/s.
Find the speed of the golf ball immediately
after impact.
9.5
Please show the way
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.
Momentum is defined as the product of an object's mass and velocity, and it is a vector quantity. In this problem, we are given the masses and velocities of the golf club head and the golf ball before and after the collision.
Let's denote the initial velocity of the golf club head as v1_club, the mass of the golf club head as m1_club, the initial velocity of the golf ball as v1_ball, the mass of the golf ball as m1_ball, the final velocity of the golf club head as v2_club, and the final velocity of the golf ball as v2_ball.
According to the conservation of momentum, we have:
(m1_club * v1_club) + (m1_ball * v1_ball) = (m1_club * v2_club) + (m1_ball * v2_ball)
Plugging in the given values:
(212 g * 29.8 m/s) + (45.9 g * 0 m/s) = (212 g * 21.8 m/s) + (45.9 g * v2_ball)
Now, let's solve for v2_ball.
(212 g * 29.8 m/s) = (212 g * 21.8 m/s) + (45.9 g * v2_ball)
(212 g * 29.8 m/s) - (212 g * 21.8 m/s) = 45.9 g * v2_ball
Simplifying the equation:
(212 - 212) g * 29.8 m/s = 45.9 g * v2_ball
0 = 45.9 g * v2_ball
Since the mass of the golf ball, m1_ball, is non-zero, the only way for the equation to be true is if v2_ball is equal to 0 m/s. Therefore, the speed of the golf ball immediately after impact is 0 m/s.