A roller coaster ride at an amusement park lifts a car of mass \( 700 \, \text{kg} \) to point \( A \) at a height of \( 90 \, \text{m} \) above the lowest point on the track, as shown above. The car starts from rest at \( A \), rolls with negligible friction down the incline and follows the track around a loop of radius \( 20 \, \text{m} \). Point \( B \), the highest point on the loop, is at a height of \( 50 \, \text{m} \) above the lowest point on the track.(a) Calculate the maximum speed of the car, when it reaches the very bottom of the track . , (b) Calculate the speed \( v_B \). , (c) Draw and label the vectors to represent the forces acting on the car when it is upside down at point \( B \). , (d) Calculate the magnitude of all the forces identified in (c). If the shape or size of the loop were modified, what changes (if any) would be needed to ensure that the roller coaster still has the same speed at the top of the loop as calculated in part (b)? Justify your answer. :

\(\boxed{v_{\text{max}} = 42 \, \text{m/s}}\) \(\boxed{v_B = 28 \, \text{m/s}}\) Forces at Point B: Normal Force (\( F_N \)): Points towards the center of the loop. Gravitational Force (\( F_g \)): Points downwards (also towards the center when car is at the top of the loop).\(\boxed{F_N = 20580 \, \text{N}}\), \(\boxed{F_g = 6860 \, \text{N}}\) If the loop’s shape or radius is modified, the drop height should be adjusted accordingly to ensure energy conservation and keep \( v_B \) unchanged. This is supported by the equation derived in part B: \(v_B = \sqrt{2g (h_A – h_B)}\) which shows that \( v_B \) depends only on the height difference of the rollercoaster.

A roller coaster ride at an amusement park lifts a car of mass 700 kg to point A at a height of 90 m above the lowest point on the track, as shown above. The car starts from rest at A, rolls with negligible friction down the incline and follows the track around a loop of radius 20 m. Point B, the highest point on the loop, is at a height of 50 m above the lowest point on the track. • Nerd Notes

AP Physics

Unit 4 - Energy

FRQ

Mathematical

Advanced

(a) Calculate the maximum speed of the car at the lowest point of the track.

Step	Derivation/Formula	Reasoning
1	\[ PE_A = mgh_A \]	Calculate the potential energy at point \( A \) where \( h_A = 90 \, \text{m} \).
2	\[ KE_P = \frac{1}{2}mv_{\text{max}}^2 \]	Calculate the kinetic energy at point \( P \). The speed is at maximum here, so \(\Delta h = 0\).
3	\[ PE_A = KE_P \]	Use the conservation of energy: all potential energy at \( A \) converts into kinetic energy at \( P \).
4	\[ mgh_A = \frac{1}{2}mv_{\text{max}}^2 \]	Set the potential energy equal to the kinetic energy. The mass \( m \) cancels out.
5	\[ v_{\text{max}} = \sqrt{2gh_A} \]	Solve for \( v_{\text{max}} \) by rearranging and simplifying the equation.
6	\[ v_{\text{max}} = \sqrt{2 \times 9.8 \times 90} \]	Substitute \( g = 9.8 \, \text{m/s}^2 \) and \( h_A = 90 \, \text{m} \) into the equation.
7	\[\boxed{v_{\text{max}} = 42 \, \text{m/s}}\]	Calculate the final value of the maximum velocity \( v_{\text{max}} \).

(b) Calculate the speed \( v_B \) at the top of the loop.

Step	Derivation/Formula	Reasoning
1	\[ PE_B = mgh_B \]	Calculate the potential energy at point \( B \) where \( h_B = 50 \, \text{m} \).
2	\[ KE_B = \frac{1}{2}mv_B^2 \]	Calculate the kinetic energy at point \( B \).
3	\[ PE_A = PE_B + KE_B \]	Apply conservation of energy between points \( A \) and \( B \).
4	\[ mgh_A = mgh_B + \frac{1}{2}mv_B^2 \]	Set the potential and kinetic energy sum at \( B \) equal to the potential energy at \( A \). Mass \( m \) cancels out.
5	\[ gh_A = gh_B + \frac{1}{2}v_B^2 \]	Mass cancels out; simplify the equation for \( v_B \).
6	\[ v_B^2 = 2g(h_A – h_B) \]	Rearrange to solve for the speed \( v_B \).
7	\[ v_B = \sqrt{2 \times 9.8 \times (90 – 50)} \]	Substitute values \( g = 9.8 \, \text{m/s}^2 \), \( h_A = 90 \, \text{m} \), and \( h_B = 50 \, \text{m} \) into the equation.
8	\[\boxed{v_B = 28 \, \text{m/s}}\]	Calculate the speed \( v_B \) at the top of the loop.

(c) Forces at Point B:

– Normal Force (\( F_N \)): Points towards the center of the loop.
– Gravitational Force (\( F_g \)): Points downwards (also towards the center when car is at the top of the loop).

(d) Calculate the magnitude of all forces and explain loop modification.

Step	Derivation/Formula	Reasoning
1	\[ F_N + F_g = \frac{mv_B^2}{r} \]	The centripetal force required at point \( B \). Both forces point towards the center.
2	\[ F_g = mg \]	Substitute gravitational force where \( m = 700 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
3	\[ F_N + mg = \frac{mv_B^2}{r} \]	Substitute the gravitational force into the centripetal force equation.
4	\[ F_N = \frac{mv_B^2}{r} – mg \]	Rearrange to solve for the normal force \( F_N \).
5	\[ F_N = \frac{700 \times 28^2}{20} – 700 \times 9.8 \]	Substitute known values: \( m = 700 \, \text{kg} \), \( v_B = 28 \, \text{m/s} \), \( r = 20 \, \text{m} \).
6	\[\boxed{F_N = 20580 \, \text{N}}\]	Calculate the normal force at the top of the loop.

Loop Modification:

If the loop’s shape or radius is modified, the drop height should be adjusted accordingly to ensure energy conservation and keep \( v_B \) unchanged. This is supported by the equation derived in part B: \(v_B = \sqrt{2g (h_A – h_B)}\) which shows that \( v_B \) depends only on the height difference of the rollercoaster.

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Circular Motion, Energy

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\(\boxed{v_{\text{max}} = 42 \, \text{m/s}}\)
\(\boxed{v_B = 28 \, \text{m/s}}\)
Forces at Point B: Normal Force (\( F_N \)): Points towards the center of the loop. Gravitational Force (\( F_g \)): Points
downwards (also towards the center when car is at the top of the loop).\(\boxed{F_N = 20580 \, \text{N}}\), \(\boxed{F_g = 6860 \, \text{N}}\)
1. If the loop’s shape or radius is modified, the drop height should be adjusted accordingly to ensure energy conservation and keep \( v_B \) unchanged. This is supported by the equation derived in part B: \(v_B = \sqrt{2g (h_A – h_B)}\) which shows that \( v_B \) depends only on the height difference of the rollercoaster.

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

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