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A mass 100 g strikes the wall with speed 5 m/s at an angle of 60 degree with a wall and rebounds with same speed. If the contact time is 2 * 1/1000 sec. , what is force applied by wall.

Hello Ankit,

Note that the angle between the rebound and the wall will also be 60 degrees. The component parallel to the wall of the original momentum continues unchanged. The components perpendicular to the wall of the original momentum and the final momentum are equal and opposite. Let

Vi = 5 m/s.

Since the force applied by the wall is away from the wall, I will give the component perpendicular to the wall of the final momentum a positive sign so that the wall's force will also have a positive sign.

Therefore:

the component perpendicular to the wall of the original momentum = -m*Vi*sin60

(the minus sign indicates that it is toward the wall) and

the component perpendicular to the wall of the final momentum = m*Vi*sin60

So these 2 components are equal and opposite. The wall gives the mass an impulse. Impulse is equal to the change in momentum. The only change in this case was in the perpendicular component. Subtracting the initial perpendicular component from the final perpendicular component, we find that

the change in momentum = m*Vi*sin60 - (-m*Vi*sin60) = 2*m*Vi*sin60

I said that Impulse = change in momentum. Also, the impulse = the product of the applied force and the time of contact. Therefore

2*m*Vi*sin60 = F*t

Plugging our data in and solving for F,

2*0.1 kg*(5 m/s)*sin60 = F*0.002 s

F = 2*0.1 kg*(5 m/s)*sin60 / 0.002 s = 433 kg.m/s^2 = 433 N

I hope this helps,

Steve

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