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Question: A small sphere of mass m=1kg is moving with a velocity (\[4\widehat{i}-\widehat{j}\])m/s. It hits a ...

A small sphere of mass m=1kg is moving with a velocity (4i^−j^4\widehat{i}-\widehat{j})m/s. It hits a fixed smooth wall and rebounds with velocity ( 4i^+3j^4\widehat{i}+3\widehat{j} )m/s. The coefficient of restitution between the sphere and the wall is n/16. Find the value of n.

Explanation

Solution

The impulse comes into play when a very large force acts on an object and that too for a very short time. It quantifies the overall effect of a force acting over time. According to Newton’s third law, forces always occur in pairs. We are given with the velocity vectors and coefficient of restitution.

Complete step by step answer:
Initial velocity=4i^−j^4\widehat{i}-\widehat{j}
Final velocity= 4i^+3j^4\widehat{i}+3\widehat{j}
Coefficient of restitution, e= n16\dfrac{n}{16}
During the collision, the component of velocity parallel to the wall remains unchanged. Let p^\widehat{p}be the normal unit vector such that p^=ξi^+ζj^\widehat{p}=\xi \widehat{i}+\zeta \widehat{j}and ξ2+ζ2=1{{\xi }^{2}}+{{\zeta }^{2}}=1
v→−u→=((v→−u→)p^)p^\overrightarrow{v}-\overrightarrow{u}=((\overrightarrow{v}-\overrightarrow{u})\widehat{p})\widehat{p}
We are given the initial and final velocity vectors as 4i^−j^4\widehat{i}-\widehat{j}and 4i^+3j^4\widehat{i}+3\widehat{j}

& \therefore \xi =\dfrac{-3}{5} \\\ & \zeta =\dfrac{4}{5} \\\ \end{aligned}$$ Thus, our unit vector comes out to be =$$\widehat{p}=\dfrac{-3}{5}\widehat{i}+\dfrac{4}{5}\widehat{j}$$ We know that, Coefficient of restitution, $$e=-\dfrac{v\widehat{p}}{u\widehat{p}}=\dfrac{-3+12}{-12-4}=\dfrac{9}{16}$$ But e= $$\dfrac{n}{16}$$ So, $$\dfrac{n}{16}=\dfrac{9}{16}$$ n= 9. **Thus, the value of n comes out to be 9.** **Additional Information:** 'e' is a consequence of Newton's Experimental Law of Impact. the coefficient of restitution ​has no unit; it is a dimensionless ratio. It is the ratio of the relative velocity of an object after collision to the relative velocity of the object before collision. It lies between 0 to 1. **Note:** We have introduced a normal vector which is perpendicular to the wall because, during the collision, the component of the velocity parallel to the wall remains unchanged. During the Collision the components of the velocity along the wall cancel out each other.