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# 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.

Last updated date: 20th Jun 2024
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Hint: 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:
\begin{align} & J=\Delta P \\ & \Rightarrow J=M{v}'+mV \\ \end{align}

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{align} & 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{align}
Substituting the given values in above equation we get:
\begin{align} & 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{align}
Therefore, we can say from the result that the friction between the ring and the ground is to the left.