Question

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

Answer 1

Answer:

Here initial velocity is u and final velocity is v

u is making angle θ and v is making angle ϕ with x−axis

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

⟹x−component of initial velocity ,ucosθ will remain unchaged

⟹V  

x

​  

=ucosθ

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(usinθ)  

Here it must be noted that

cefficient of restitution is used along common normal

Considering ball,

J=mv  

f

​  

−mv  

i

​  

 , where v  

f

​  

 and v  

i

​  

 are final and initial velocities of ball along common normal,

V  

i

​  

=−usinθ(−y−direction)

⟹J=meusinθ−m(−usinθ)=musinθ(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

 

Explanation: