Question

a mass on moving with velocity u hits a surface at an angle theta with normal to surface. how much impulse does the surface exert

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

⟹Vx=ucosθ

for Vy 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)

Vy=e(usinθ)  

Here it must be noted that

 cefficient of restitution is used along common normal

Considering ball,

J=mvf−mvi , where vf and vi are final and initial velocities of ball along common normal,

Vi=−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

Answer 2

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 directio