Question

A ball of mass $m$ moving with speed $u$ collides with a smooth horizontal surface at angle $\theta$ with it as shown in figure. The magnitude of impulse imparted to surface by ball is [Coefficient of restitution of collision is e]

• A. $mu(1+e)\cos$$\theta$
• B. $mu(1-e)\sin$$\theta$
• C. $mu(1-e)\cos$$\theta$
• D. $mu(1+e)\sin$$\theta$

Answer 1

The correct option is D $mu(1+e)\sin$$\theta$

Here initial velocity is $u$ and final velocity is $v$

$u$ is making angle $\theta$ and $v$ is making angle $\varphi$ with $x-axis$

The impulse $J$ acting on the ball is along $common$ $normal$ $(y-axis)$

$⟹x-component$ of initial velocity ,$u\cos \theta$ will remain unchaged

$⟹V_x=u\cos \theta$

for $V_y$ we will use coefficient of restitution

we know that, (velocity of separation )$=e$ (velocity of approach),

(in the above expression magnitude of velocities are used)

$V_y=e\left(u\sin \theta \right)$

Here it must be noted that

cefficient of restitution is used along common normal

Considering ball,

$J=m{v}_f-m{v}_i$ , where $v_f$ and $v_i$ are final and initial velocities of ball along common normal,

$V_i=-u\sin \theta (-y-direction)$

$⟹J=meu\sin \theta -m(-u\sin \theta )=mu\sin \theta (1+e)$

$J=$ impulse imparted by surface to the balland by newton's third law we know that this much impulse will act on surface in opposite direction

$answer=D$