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almond37 [142]
3 years ago
12

Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet

ely and throws Sheila forward onto the pavement. If there is no net force on the system, what is Sheila's velocity immediately after she is thrown from the bike? 1.8 m/s 2.0 m/s 5.0 m/s 5.7 m/s
Physics
1 answer:
Vikki [24]3 years ago
6 0

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

<u>v₂ = 5.7 m/s </u>

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          μ = m / l

iinitial density

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final density

          μ = 2m₀ / lo

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we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

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             v = 1 /√5  v₀

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