High-speed stroboscopic photographs show
that the head of a 192 g golf club is traveling
at 44.4 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 28.7 m/s.
Find the speed of the golf ball immediately
after impact.
Answer in units of m/s.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Therefore, the momentum of the club head before the collision is (192 g) * (44.4 m/s) and the momentum of the golf ball after the collision is (45.9 g) * v (where v is the velocity of the ball immediately after impact).
Since the collision is elastic (no energy is lost), we can equate the momenta before and after the collision:
(192 g) * (44.4 m/s) = (45.9 g) * v
Now, we can solve for v by rearranging the equation:
v = (192 g * 44.4 m/s) / (45.9 g)
Note: We need to convert the masses to kilograms to ensure consistency with SI units. 1 g = 0.001 kg.
v = (192 kg * 0.0444 m/s) / (0.0459 kg)
v ≈ 186.93 m/s
Therefore, the speed of the golf ball immediately after impact is approximately 186.93 m/s.
M1*V1 + M2*V2 = M1*28.7 + M2*V.
0.192*44.4 + 0.459*0 = 0.192*28.7 + 0.459V.
V = ?