In bowling the ball rolls on the floor with initial speed of 15 m/s and sweeps the horizontal distance of 125 m before it comes to rest. F ind the coefficient of friction between the ball and floor?
Answers
Answered by
2
Answer:
Velocity of approach in the given case is the normal component of velocity.
Hence, vn=vcosθ
By definition of coefficient of restitution, velocity of separation will be,
vn′=evn=evcosθ
Tangential component of velocity will not change.
vt′=vt=vsinθ
Speed of reflected ball is:
v′=vn2+vt2
Angle with normal is given by:
θ′=tan−1vn′vt′
θ′=tan−1(evcosθvsinθ)
θ′=tan−1(etanθ)
Similar questions