Question

The point B lies on a smooth plane inclined at ${30}^{\circ }$ to the horizontal. A particle of mass $\frac{1}{7}\text{kg}$ is dropped from a point A which lies $10\text{m}$ vertically above B. The particle rebounds from the plane in the direction BC with speed $v\text{m/s}$ at an angle of ${45}^{\circ }$ to the plane. Find the impulse exerted by the plane on the particle (in $\text{Ns}$)

• A. $1+\sqrt{3}$
• B. $1-\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $2-2\sqrt{3}$

Answer 1

The correct option is A $1+\sqrt{3}$

As the plane is smooth, the impulse must be perpendicular to the plane.
Velocity of particle just before the collision is $u=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}=14\text{m/s}$



Since A is vertically above B, $\theta ={30}^{\circ }$
Let $v$ be the velocity of particle after collision.
No momentum exchange parallel to the plane.
Hence, $14\sin {30}^{\circ }=v\cos {45}^{\circ }$
or $v=7\sqrt{2}\text{m/s}$

Impulse $|J|=[mv\sin {45}^{\circ }-m(-u\cos {30}^{\circ }\left)\right]$
$|J|=\frac{1}{7}[7\sqrt{2}\times \frac{1}{\sqrt{2}}+14\times \frac{\sqrt{3}}{2}]$
$|J|=(1+\sqrt{3})\text{Ns}$