$\textbf{Problem Statement}$ $\mathbf{Attempt}$ Assume that the sphere is moving along the X axis.

Let the particle strike the sphere such that the radius vector of point of collision makes an angle $\theta$ with the horizontal plane (the xy plane) and its projection on xy plane make an angle $\phi$ with X axis. For the collision, the velocity of the sphere in the xy plane along the line of impact is $v \cos \phi$. We have now reduced a 3D oblique collision into a 2D one. This velocity's projection on the line of impact is $\boxed{u=v \cos \phi \cos \theta}$. Thus the collision takes place along the radius vector whose direction is $\hat{r}=\cos \phi \cos \theta \hat{i} - \sin \phi \cos \theta \hat{j} + \sin \theta \hat{k}$ ow since the sphere is large, final velocity of particle along $\hat{r}$ is $2u$, and along the X direction, the total change in momentum of particle is $\Delta P_x= mu_x=\boxed{2mv \cos^2 \phi \cos^2 \theta}$ Now the volume swept out by the region in time $\Delta t$ is $dV= vR^2 \Delta t \cos \phi \cos \theta { ~d \phi }{ ~d \theta }$ and the number of particles hitting is $n \cdot dV$, thus the total force acting on the sphere along X axis is $F= 2mv^2nR^2 \color{#3D99F6}{\underbrace{\color{#D61F06}{\displaystyle \int_{-\pi /2}^{\pi /2} \int_{-\pi /2}^{\pi /2} \cos^3 \phi \cos^3 \theta ~d \phi ~d \theta}}_{I}}.$ We can integrate over $\phi$ for a particular $\theta$ and then integrate over $\theta$. So $\boxed{F=\dfrac{32}{9}mv^2nR^2}$. $\textbf{Official Solution}$ I don't quite get how the theta $(\theta)$ is taken here, (from which axis or plane) and also the answer obtained is different from mine by a numerical factor ( 32/9 Vs $\pi$). Please help on this.

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TopNewest@Gordon Chan @Steven Chase @Mark Hennings sir please help.

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@Md Zuhair please help.

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Sorry bro... I have not checked Brilliant from 4 months now! Sorry to see this late...

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Its ok. Any inputs on the problem?

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