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emmasim [6.3K]
3 years ago
14

Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s

away from the wall. if the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball?
Physics
1 answer:
Inga [223]3 years ago
8 0

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

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Explanation:

The range <em>R</em> of a projectile is given the equation

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The maximum range is achieved when \theta = 45° so our equation reduces to

R_{max} = \dfrac{v_0^2}{g}

We can solve for the initial velocity v_0 as follows:

v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}

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v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}

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Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

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