Question

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?​

Answer 1

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)