Question

A thin ring of mass 2kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1m/s. A small ball of mass 0.1kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75m and goes vertically up with velocity 10m/s, immediately after the collision.

Answer 1

Here, we will use the basic concept of impulse-momentum theorem and then we will use the equation to find the final velocity of the particle. We will further study how friction affects the motion of an object or particle. This will give us the required result.
Formula used:

& J=\Delta P \\\ & \Rightarrow J=M{v}'+mV \\\ \end{aligned}$$ **Complete answer:** Since, the impulse momentum theorem can be stated as: a collision, when an object experiences a force for a given amount of time (small duration) that results in its mass undergoing a change in velocity. Here the momentum of the two collision particles changes. From the impulse momentum theorem, the impulse imparted by the ball is J and v is the initial velocity and v’ is the final velocity. $$\begin{aligned} & J=\Delta P \\\ & \Rightarrow J=M{v}'-m\left( -V \right) \\\ & \Rightarrow J=M{v}'+mV \\\ & \Rightarrow J\times 0.25=M{{R}^{2}}\left( {{\omega }_{f}} \right)-M{{R}^{2}}\left( \dfrac{v}{R} \right) \\\ & \Rightarrow J=4M{{R}^{2}}\left( {{\omega }_{f}} \right)+4M{{R}^{2}}\left( \dfrac{v}{R} \right) \\\ & \Rightarrow 4M{{R}^{2}}{{\omega }_{f}}+4MRv=M{v}'+Mv \\\ \end{aligned}$$ Substituting the given values in above equation we get: $$\begin{aligned} & 4\times \dfrac{5}{10}\times \dfrac{8}{10}{{\omega }_{f}}+4\times \dfrac{8}{10}\times 1={v}'+1 \\\ & \Rightarrow {v}'={{\omega }_{f}}+1 \\\ & \because 2R{{\omega }_{f}}+1={v}' \\\ & \therefore {v}'\rangle {{\omega }_{f}}f \\\ \end{aligned}$$ Therefore, we can say from the result that the friction between the ring and the ground is to the left. **Additional information:** As we know that, the friction force is defined as the force exerted by a surface as an object moves across it or makes an effort to move across it. There are mainly three types of friction forces: sliding friction, static friction and the last one rolling friction. Static friction comes into action on objects when they are resting on a surface. For example, if we are hiking in the woods, there is a presence of static friction between our shoes and the trail each time we put down our foot. Without this friction, our feet would slip out and thereby, making it difficult to walk. Sliding friction is friction that comes in action on objects when they are sliding over another surface. As we know that the sliding friction is weaker than the static friction. That is the reason that it is easier to slide a piece of furniture over the floor after we start it moving than it is to get it moving in the first place. The last one is rolling friction is friction that comes in action on objects when they are rolling over another surface. Rolling friction is much weaker than sliding friction and static friction. This is the reason that most forms of ground transportation use wheels, including bicycles, cars, 4-wheelers, roller skates, scooters, and skateboards, as these wheels make it easier to travel with facing less frictional force. **Note:** Frictional force is a force that opposes the motion of any object. There are many practical examples of frictional force. We should also remember that various kinds of lubricants are used to overcome the frictional force like- grease, oil etc.