Question

A uniform ball collides with a smooth surface and rebounds . during collision impulsive force F acting on ball varies with time as show in figure . coefficient of restitution is​

Answer 1

Answer:

mu ( 1+e) sine O

Explanation:

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

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