Answer:
v1=21.81m/s
Explanation:
<em>When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 44 m/s. Neglect the mass of the club handle and determine the speed (in m/s) of the golf ball just after impact.</em>
According to the law of conservation of momentum, if the net external force on a system is zero, then the linear momentum of the system is conserved.
During collision of two particles, the external force on the system of two colliding particles is zero (only internal force acts between the colliding particles), therefore, the momentum is conserved during the collision.
Answer and Explanation:
Given :
head of the golf club=160g
velocity of the golf club=50 m/s
golf ball mass=46g
velocity=om/s
m1u1+m2u2=m1v1+m2v2.........................................1
160*50 +46*0=160*44+46*v1
8000=7040+46v1
960=46v1
v1=960/46
v1=21.81m/s
