Question

A bar of m = 1.00 kg and length l = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted so that the bar is free to rotate in the vertical plane. a small m = 0.10 kg is moving on the same horizontal surface with 5.00 ms^-1 speed on its path which is perpendicular to the bar. Its hits the bar at a distance \(\frac{L}{2}\)  from the pivoted end and return back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity ω. Which of the following statement is correct? 

• A. ω = 3.75 rad/s and v = 7.30 m/s
• B. ω = 6.80 rad/s and v = 9.10 m/s
• C. ω = 3.75 rad/s and v = 10.0 m/s
• D. ω = 6.98 rad/s and v = 4.30 m/s

Answer 1

The Correct Option is D

For the (rod+ball) system:
(i)The angular momentum remains constant only along with hinge axis.
the hinge axis = L_i=L_f
0+(0.1)(5)(0.1)=(\(\frac{1(0.2)^2}{3}\))\(\omega\)\(- (0.1)(v)(0.1)\)
= \(\frac{4\omega}{3}-v=5\,\,\,.........(1)\)
(ii) \(e=\frac{\omega(0.1)+v}{5} where\,\, e=1\)
\(\frac{\omega}{10}+v=5=6.98\,\, rad/sec\)
v=4.30 m/sec.

Answer 2

A bar, weighing 1 kilogram and measuring 0.2 meters in length, is attached at one end, called point P. A small object with a mass of 0.1 kilograms moves sideways towards the bar at a speed of 5 meters per second. When it hits the bar, it bounces off elastically, and the bar starts to spin. The object's new speed after the collision is called v, and the bar starts spinning with an angular speed called w.

Before the collision, the combined system of the bar and the object has its spin momentum going in a clockwise direction. We can represent this initial spin momentum as mu.

After the collision, the total spin momentum of the system changes. It's now represented as the bar's rotational inertia, 𝑀𝐿², times its spin speed, w, minus the object's mass times its speed after the collision, v, times the distance between the point of impact and the pivot point, L.

During the collision, the system's total spin momentum remains the same. This principle helps us find a relationship between the system's spin momentum before and after the collision:

\(2𝑀𝐿𝑤=3𝑚(𝑢+𝑣)\)

This equation tells us how the spin momentum is conserved in the system, connecting the initial and final states of the bar and the object.

The system's total kinetic energy stays the same in an elastic collision. Before the collision, the kinetic energy of the system is \(\frac{1}{2}𝑚𝑢^2\). After the collision, the total kinetic energy is the kinetic energy of the object plus the rotational kinetic energy of the bar:

\(\frac{1}{2}𝑚𝑣^2+\frac{1}{2}𝑀𝐿^2𝑤^2\)

We can use the principle of conservation of kinetic energy to equate the initial and final kinetic energies:
\(𝑚𝑢^2=3𝑚(𝑢^2−𝑣^2)\)

By solving these equations, we find that after the collision, the object's speed is 4.3 meters per second, and the bar's spin speed is 6.98 radians per second.