Physical Science question

Question

A 2400 kg roller coaster car travels 4.7 m/s over a 16 M Hill, then swoops down and climbs another 9.2 M hill. How fast is it going at the top of the 9.2 M Hill?

Answer 1

The question is asking us for final velocity. This can be done as a kinematics questions or as an energy question. 2400kg is extra information because the final velocity is mass independent.

 

 

v_f²=v₀²+2a(Δx)

v_f=√v_o²+2aΔx)

√(4.7²+2(-9.81)(9.2-16)= 12.5m/s

 

 

Assuming no friction or wind resistance, there is no change of total energy. It is a conversion of potential energy to kinetic energy.

 

ΔE=ΔKE+ΔPE

0= (1/2mvf² - 1/2mv₀²) + (mgh_f - mgh₀)

0=1/2m(v_f² - v₀²) + mg(h_f-h₀)

mg(hf-h0)=1/2m(v_f² - v₀²) 

Masses cancel (mass independent)

g(h_f-h₀)=1/2(v_f² - v₀²)

 

Isolating v_f

v_f=√v₀²+2aΔh)

√(4.7²+2(-9.81)(9.2-16)= 12.5m/s

Answer 2

Solve by energy conservation.

 

mv₁²/2 + mgh₁ = mv₂²/2 + mgh₂

 

v₁²/2 + gh₁ = v₂²/2 + gh₂

 

v₂² = 2[v₁²/2 + g(h₁ - h₂)] = 2[4.7²/2 + 9.8(16 - 9.2)] m²/s² = 155.37 m²/s²

 

v₂ = √155.37 m/s ≅ 12.5 m/s

 

It goes 12.5 m/s at the top of the 9.2 m hill.