A ball of mass m is attached to a cord of length L, pivoted at point O, as shown in the figure. The ball is released from rest at point A, swings down and makes an inelastic collision with a block of mass 2m kept on a rough horizontal floor. The coefficient of restitution of collision is $e = 2 / 3$ and coefficient of friction between block and surface is $μ$ . After collision, the ball comes momentarily to rest at C when cord makes an angle of $θ$ with the vertical and block moves a distance of 3L/2 on rough horizontal floor before stopping. The value of $μ$ and $θ$ are, respectively,

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2243, cos − 1 80243

281, cos − 1 80243

5081, cos − 1 8081

50243, cos − 1 8081

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Velocity of the ball just before collision is $v 0 = 2 g L$

Before collision:
Question Image
After collision:

Applying momentum conservation along horizonal direction (because momentum
is conserved in collision along the line of impact), we get
$m v 0 = − m v 1 + 2 m v 2$

Applying coefficient of restitution equation, we get
$e = 23 = v 1 + v 2 v 0$
Solving the above two equation, we get, $v_{1} = \frac{v_{0}}{9}, v_{2} = \frac{5 v_{0}}{9}$
As the block moves a distance of $3 L / 2$ before coming to rest, so from work-energy theorem,
$0 = \frac{1}{2} (2 m) v_{2}^{2} - μ \times 2 m g \frac{3 L}{2} \Rightarrow v_{2}^{2} = 3 μ g L$
$\Rightarrow \frac{25}{81} \times 2 g L = 3 μ g L \Rightarrow μ = \frac{50}{243}$
For ball, KE is converted to gravitational potential energy after collision, so $0 − m v 12 2 = − m g × L (1 − cos θ)$
$\Rightarrow \cos θ = \frac{80}{81}$